First, we prove Rolle's Theorem: if $f : [a, b] \to \mathbb{R}$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists $c \in (a, b)$ with $f'(c) = 0$.

By the Extreme Value Theorem (which follows from Heine-Borel compactness of $[a, b]$), $f$ attains its maximum $M$ and minimum $m$ on $[a, b]$.

Case 1: If $M = m$, then $f$ is constant, so $f'(x) = 0$ everywhere. Any $c \in (a, b)$ works.

Case 2: If $M > m$, then at least one of $M$ or $m$ differs from $f(a) = f(b)$. Suppose $M > f(a)$. Then $M$ is attained at some $c \in (a, b)$ (not at the endpoints, since $f(a) = f(b) < M$).

At this interior maximum $c$, for small $h > 0$:

$\frac{f(c+h) - f(c)}{h} \leq 0 \quad \text{(since } f(c+h) \leq f(c)\text{)}.$

Taking $h \to 0^+$, $f'(c) \leq 0$. Similarly, for $h < 0$ small,

$\frac{f(c+h) - f(c)}{h} \geq 0 \quad \text{(since } f(c+h) \leq f(c) \text{ and } h < 0\text{)}.$

Taking $h \to 0^-$, $f'(c) \geq 0$. Thus $f'(c) = 0$.

Now prove the MVT. Define

$g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a).$

Then $g$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $g(a) = g(b) = 0$. By Rolle's Theorem, there exists $c \in (a, b)$ with $g'(c) = 0$.

Computing $g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}$, we get

$g'(c) = 0 \implies f'(c) = \frac{f(b) - f(a)}{b - a}.$