Are there alternative proofs of the general Taylor-series expansion theorem for real functions?
It's simple to discover Taylor series. Let's start with $$ \tag{1}f(x) = f(a) + \int_a^x f'(s) \, ds, $$ which of course is just the fundamental theorem of calculus. Now if we are feeling playful we might note (again by FTC) that $f'(s) = f'(a) + \int_a^s f''(t) \, dt$. Plugging this into (1), we find that \begin{align} f(x) &= f(a) + \int_a^x f'(a) + \int_a^s f''(t) \, dt \,ds \\ \tag{2}&= f(a) + f'(a)(x - a) + \underbrace{\int_a^x \int_a^s f''(t) \, dt \, ds}_{\text{remainder}}. \end{align} We can keep going like this for as long as we want. The next step is to note that $f''(t) = f''(a) + \int_a^t f'''(u) \, du$. Plugging this into (2), we find that \begin{align} f(x) &= f(a) + f'(a) (x - a) + \int_a^x \int_a^s f''(a) + \int_a^t f'''(u) \, du \, dt \, ds \\ &= f(a) + f'(a)(x - a) + \int_a^x f''(a)(s - a) + \int_a^s \int_a^t f'''(u) \, du \, dt \, ds \\ &= f(a) + f'(a)(x - a) + f''(a) \frac{(x-a)^2}{2} + \underbrace{\int_a^x \int_a^s \int_a^t f'''(u) \, du \, dt \, ds}_{\text{remainder}}. \end{align} You see the pattern. So we have discovered the Taylor polynomial approximation to $f(x)$, and we have a formula for the remainder.
By the way, if $| f'''(u) | \leq M$ for all $u \in [a,x]$, then the remainder $R(x)$ satisfies \begin{align} | R(x) | &\leq \int_a^x \int_a^s \int_a^t | f'''(u) | \, du \, dt \, ds \\ &\leq \int_a^x \int_a^s \int_a^t M \, du \, dt \, ds \\ &= M \frac{(x-a)^3}{3!}. \end{align} You see what the bound on the remainder will be for higher order Taylor series approximations. So we see that the remainder will be small if $x$ is close to $a$.
(If $f$ is sine or cosine, we can take $M = 1$. If $f$ is the exponential function, we can take $M = e^x$.)