Proof of Rolle's Theorem in One Dimension

By Extreme Value Theorem on a Compact Interval, there exist points $x_{\min},x_{\max}\in[a,b]$ such that

If $f(x_{\min})=f(x_{\max})$, then every value of $f$ on $[a,b]$ is equal to this common value, so $f$ is constant on $[a,b]$. Since $a<b$, choose any $c\in(a,b)$. For every $h$ with $c+h\in[a,b]$ and $h\ne 0$ one has

and therefore $f'(c)=0$.

Assume now that $f(x_{\min})<f(x_{\max})$. Because $f(a)=f(b)$, the points $a$ and $b$ cannot both realize the maximum and the minimum. More precisely, if $x_{\max}\in\{a,b\}$, then $f(x_{\max})=f(a)=f(b)$, so the strict inequality $f(x_{\min})<f(x_{\max})$ forces $x_{\min}\notin\{a,b\}$, hence $x_{\min}\in(a,b)$. Similarly, if $x_{\min}\in\{a,b\}$, then $x_{\max}\in(a,b)$. Thus at least one of $x_{\min}$ or $x_{\max}$ lies in $(a,b)$. At such a point $f$ has a local extremum, so Fermat Stationary Point Criterion yields a point $c\in(a,b)$ with $f'(c)=0$.