Edit: this question reflects some sloppy thinking about derivatives, now corrected. I left the question untouched in case others may benefit. The point of clarification is that if a derivative exists at a point, which it does by hypothesis in Rolle's theorem, then its existence demands the left and right derivatives be equal, which provides justification for the theorem's conclusion.
I'm looking through George Simmons's Calculus text, in particular Appendix A.4 where he discusses Rolle's Theorem building towards the MVT proof. The proof is just like the "generalized proof" on Wikipedia: given $f(x)$ continuous on $[a,b]$ and differentiable on $(a,b)$, with $f(a)=f(b)$, and presuming $f(x)$ is non-constant, examine $c \in (a,b)$ where $f(c)$ is a maximum of $f(x)$ over $[a,b]$, you find that $\lim_{x \to c^+} f'(x) \leq 0$ and $\lim_{x \to c^-} f'(x) \geq 0$. Hence if the limits agree from the right and the left, then $f'(c)=0$, the point of the theorem.
My issue is with the final sentence. If I can assume that $f'(x)$ is continuous on $(a,b)$, by the IVT $f'(x)$ must cross zero at $c$ and the conclusion is true. But how am I assured of this continuity? I wondered if I am guaranteed continuity in $f'(x)$ by the very definition of differentiability, which in the Simmons text is that the derivative exists at a point when it is the same when approached from the right and the left. That seems to be smuggling in a guarantee that $f'(x)$ is continuous without quite saying so, for a function differentiable on $(a,b)$ has at every point the same derivative approaching from the left or right and therefore cannot have discontinuities in its derivative.
But the Wikipedia page on differentiable functions reserves "continuously differentiable" to mean a guarantee that $f'(x)$ is continuous, clearly noting that there are cases where a function is differentiable (hence continuous) but the derivative is not continuous. All proofs I've seen of Rolle's Theorem mention only "differentiable".
My question: it seems for Rolle's Theorem to get the zero-crossing of $f'(c)=0$, the derivative must itself be continuous, but where is that established? Is it the case that saying $f(x)$ is differentiable on $(a,b)$ is where I get assured that a merely "differentiable" function becomes "continuously differentiable" in the interval under study? Or, if it is not necessary that $f'(x)$ be continuous, then how are we guaranteed that $f'(c)=0$?
What you describe as the argument in Wikipedia is not accurate.
The generalization does not make assumptions about the limits of the derivatives; it does not even talk about limits of the derivatives. The generalization talks about the left and right derivatives.
Recall that the derivative is defined to be $$f'(c) = \lim_{h\to 0}\frac{f(c+h)-f(c)}{h}.$$ The left and right derivatives, which Wikipedia denotes $f'(c^-)$ and $f'(c^+)$, and are often denoted $f'_{-}(c)$ and $f'_{+}(c)$, are defined by $$\begin{align*} f'(c^-)&=\lim_{h\to 0^-}\frac{f(c+h)-f(c)}{h}&\text{(left derivative at }c\text{)}\\ f'(c^+)&=\lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}&\text{(right dervivative at }c\text{)}. \end{align*}$$ From the properties of limits, we have that $f'(c)$ exists if and only if $f'(c^-)$ and $f'(c^+)$ both exists, and $f'(c^-)=f'(c^+)$.
If $f(x)$ has a maximum at $c$, and we have that the one-sided derivatives exist at $c$, then $$\lim_{h\to 0^-}\frac{f(c+h)-f(c)}{h}\leq 0\text{ and }\lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}\geq 0,$$ which is not the same as $\lim_{x\to c^-}f'(x)\leq 0$ and $\lim_{x\to c^+}f'(x)\geq 0$, as you assert. In other words, we are verifying that $f'(c^-)\leq 0$ and $f'(c^+)\geq 0$.
If $f$ is differentiable at $c$, then they must be equal to each other, which is only possible of they are both $0$, in which case $f'(c)$ must also equal $0$.
That said, while the derivative of a continuous function is not necessarily continuous, nonetheless it satisfies the Intermediate Value Property. This is Darboux's Theorem: