Proof of Rolle's Theorem in One Dimension

By Extreme Value Theorem on a Compact Interval, the function $f$ attains a maximum and a minimum on $[a,b]$. Let $x_{\max},x_{\min}\in[a,b]$ be points such that $f(x_{\max})\ge f(x)\ge f(x_{\min})$ for all $x\in[a,b]$.

If $f(x_{\max})=f(x_{\min})$, then $f$ is constant on $[a,b]$, so $f'(x)=0$ for every $x\in(a,b)$. In particular, any $c\in(a,b)$ works.

Assume now that $f$ is not constant. Then either $f(x_{\max})>f(a)=f(b)$ or $f(x_{\min})<f(a)=f(b)$. In the first case $x_{\max}\in(a,b)$; in the second case $x_{\min}\in(a,b)$. Thus there exists an interior point $c\in(a,b)$ at which $f$ has a local extremum. By Fermat Stationary Point Criterion, we have $f'(c)=0$.