Will the night sky eventually be bright?
UPDATE: As Zassounotsukushi correctly points out in the comments, my original answer was wrong. I said before that objects move out across our horizon, but they don't. Sorry about that. I hope I've fixed things now.
It's best to avoid phrases like "the Universe is expanding faster than the speed of light." In general relativity, notions like distances and speeds of faraway objects become hard to define precisely, with the result that sentences like that have no clear meaning.
But if we leave that terminological point aside, your questions are perfectly well-posed and physically meaningful.
It's true that we can only see out to a finite distance, due to the Universe's finite age, and that this is an explanation of Olbers's paradox, which is the name for the old puzzle of why the night sky is dark.
As the Universe expands, more objects pop into view, since that "horizon distance" is continually getting bigger, at least in principle. In fact, though, that's a very small effect and would not lead to the night sky becoming brighter in practice.
First, we should note that we wouldn't expect to see stars popping into view as our horizon expands. The reason is that objects right near the edge of the horizon are so far away that we would see them as they were long ago, around the time of the Big Bang. In practice, we can't see all the way back to $t=0$, because the early Universe was opaque, but still, we can see back in time to long before there were discrete objects like stars. When we look back near the horizon, we see a nearly-uniform plasma.
But there's a much more important point. The further away we try to look, the more redshifted the light from a given object is. Even if we could see an object near our horizon, the radiation from it would be shifted to extremely long wavelengths, which also means that it would carry extremely little energy. In practice, this just means that things near our horizon become unobservably faint. In a practical sense, our ability to see faraway objects actually decreases with time: although in principle our horizon grows, the redshift causes any given object to become unobservably faint much faster than the rate at which new stuff is brought in across the horizon.
Lawrence Krauss has written a bunch about this stuff. The details are in this paper, and he has a Scientific American article (paywalled). Dennis Overbye wrote about this stuff in the NY Times a while back too. (If you read the pop stuff, tread carefully. Some of it seems to be saying the incorrect thing I said before, namely that things that are currently inside our horizon move outside of it. The technical article is correct, but the nontechnical ones can be misleading. That's my excuse for messing things up in my original answer, but it's not a very good excuse, because I should've known better.)
The way that you ask your question confuses the answer because you say "Will the speed of universe expansion will make the sky bright but the red shift make it invisible to our eyes", because the sky is already "bright" in certain wavelengths, particularly the centimeter cosmic background radiation (CMB).
Other than this revision, yes, the observation that the night sky is dark has been a clear argument against an infinitely old universe since long ago. The evidence for the big bang in the form of a consistently increasing red shift pretty much seals the deal for myself, regarding the fact that the universe has an age.
Furthermore, over time, you are entirely correct in the assertion that the number of observable objects will increase drastically, and quite possibly infinitely. Consider that we only see $x$ distance away which terminates at the CMB, thus limiting the number of galaxies we can see, with the furthest galaxies being the earliest evolutionary stage of galaxies. The number of "young" galaxies we can see will progressively increase as more of the veil from the CMB is pulled back through the arrival of the new light. The "young" galaxies we can see now will mature and the total number will increase. Whether or not this will increase forever is disputable since dark energy pulls space apart could prevent it but we can't claim to know exactly what the behavior of dark energy far in to the future will be.
Additions
I started thinking about the problem more and I wanted to formalize things a bit better. Take the most basic case, we'll deal with a flat Newtonian space for now. As before, take $x$ to be the distance to a certain galaxy we are current seeing. Take the present time (after the big bang) to be $t$ and that we're observing that galaxy at $t'$. It follows...
$$x=c (t-t')$$
Imagine the universe has a galaxy density of $\rho$ galaxies per unit volume. Then knowing that, we can actually write the rate $r$ at which galaxies older than $t'$ are appearing into our view. This is done knowing the surface of a sphere is $4\pi r^2$.
$$r=4 \rho c \pi x^2$$
It's fascinating to consider that in a line connecting every object in the night sky and us, there exists the entire history of the object encoded in the light waves making their way to us. One way to talk about the acceleration of the universe is to say that there is a slowdown in the rate at which we are receiving this information. We are watching the far off objects in slow motion.
If we make the obviously incorrect but useful assumption that all objects emit light at the same rate at all times, then the intensity we see will be proportional to $1/x^2$, and given some $S$ which is, say, the number of photons emitted total per unit time, then the intensity of light we receive from a given body would be $S/(4 \pi x^2)$. Multiplying this by the rate, we can get a very nice equation for $s(x)$ which is the contribution to the number of photons we receive from the differential "shell" of stars at $x$.
$$s(x) = S \rho c $$
This equation is important because it is cumulative from time at $t'$ to $t$, meaning that the objects that entered our field of vision from the "genesis" of that type of object are still contributing to the population of photons reaching us today. So the number of photons we are receiving could be said to be:
$$\int_{t'}^t S \rho c dt = S \rho c (t-t')$$
A more advanced view of the situation simply notes that the "movie" for each of these stars is being played in slow motion. We'll just define a factor for that and put it in the equation.
$$l(x) = \frac{\Delta t_{object}}{\Delta t_{Earth}}$$
I should preface this by saying that this isn't actually saying time is going slower for that object, and this isn't even the time dilation as defined by general relativity, this is the time dilation you would measure by watching a clock in a galaxy far away with a space telescope and comparing it to the local time. Yes, these two are different, and yes, I am avoiding advanced relativistic concepts by making it an accounting problem. Now the total # of photons we receive per unit time is the following.
$$\int_{t'}^t S \rho c l(x) dt$$
I won't use any calculus chain rules because there's no guarantee that $l(t)$ is any more helpful to you than $l(x)$! But I should also note that the final $x$ you get in this equation at $t$ will be meaningless. It is not the general relativity distance, it's some bastardization of that by using $c t$, which is clearly not how it actually works. Nonetheless, there is some usefulness in the above equation. We can even identify the radiative energy being received by considering the energy of the photon being proportional to it's frequency, with $E_e$ being the energy of the emitted photon and $E_o$ the observed photon.
$$\frac{E_e}{E_o} = l(x)$$
And the total energy would then be the following with $h$ the familiar plank constant.
$$E = \int_{t'}^t S h \rho c l(x)^2 dt$$
Anyway, my intent is for these to be instructive "kindergarten" equations for the subject. The bottom line is still clear from them - that the # of photons reaching us would increase linearly over time but it's less since $l(x)\le 1$. Similarly, the radiative energy reaching us would be less by even a smaller factor due to the redshift. I hope this is a clear picture.