Infinite 'hex': is a win always achievable?
No. In fact, it is possible to color the entire unit square red and blue so that there is no nonconstant continuous function $f:[0,1]\to[0,1]\times[0,1]$ whose image is entirely one color. The key observation is that the image of any such nonconstant function is a closed subset of $[0,1]\times[0,1]$ of cardinality $\mathfrak{c}$, and there are only $\mathfrak{c}$ such closed subsets. You can then construct a coloring by transfinite induction so that each such set has both a red point and a blue point.
In detail, let $(X_\alpha)_{\alpha<\mathfrak{c}}$ be an enumeration of all the closed subsets of $[0,1]\times[0,1]$ of cardinality $\mathfrak{c}$. We define sequences $(r_\alpha)_{\alpha<\mathfrak{c}}$ and $(b_\alpha)_{\alpha<\mathfrak{c}}$ by induction. Having defined $r_\beta$ and $b_\beta$ for all $\beta<\alpha$, define $r_\alpha$ and $b_\alpha$ to be two distinct points of $X_\alpha$ which are not equal to $r_\beta$ or $b_\beta$ for any $\beta<\alpha$. This is possible since we have chosen fewer than $\mathfrak{c}$ points so far, and $X_\alpha$ has cardinality $\mathfrak{c}$.
We thus obtain two disjoint sets $R=\{r_\alpha\}_{\alpha<\mathfrak{c}}$ and $B=\{b_\alpha\}_{\alpha<\mathfrak{c}}$ which each intersect every $X_\alpha$. Color all the points in $R$ red, and all the points in $B$ blue, and all the points that are in neither $R$ nor $B$ however you want. Then neither the red points nor the blue points contain any $X_\alpha$, and thus neither contains the image of a nonconstant continuous function $f:[0,1]\to[0,1]\times[0,1]$.
However, it is true that at most one of your conditions (a) and (b) are true. See this answer to an earlier question for a proof. (The proof there assumes the endpoints of the paths are the corners of the square, but a similar argument works in general. Or, you can apply a homeomorphism of the square that sends the endpoints of the paths to the corners.)
Color $(x,y)\in[0,1]^2$ red if $x=0$ or $y=1/2\cdot\sin(1/x)$. Color everything else blue. There are no paths of either color connecting its respective edges.
Note that the red path does not "reach" the line $x=0$. See also this post and the counterexample in the answers: "Intermediate Value Theorem" for curves
Okay, thinking about it more has gotten me this answer:
Consider the colouring:
$(x,0)$ is always blue
For each natural number n,
$(\frac{1}{n},y)$ is blue iff $\frac{1}{n}\leq y$, and everything else is red.