Proof of A Continuous Function with Vanishing Derivative is Constant

If $a=b$, then $[a,b]=\{x\in\mathbb{R}:a\le x\le a\}=\{a\}$, and the only value of $f$ is $f(a)$, so the conclusion holds. Assume from now on that $a<b$.

Let $x\in[a,b]$ with $a<x$; we show $f(x)=f(a)$ (for $x=a$ there is nothing to prove). Consider the restriction $f|_{[a,x]}$ of $f$ to the closed interval $[a,x]$, which satisfies $[a,x]\subseteq[a,b]$.

By claim 1 of Restriction Stability of Continuity and of the Derivative, applied with both $(X,d_X)$ and $(Y,d_Y)$ equal to the real line, $A=[a,b]$ and $B=[a,x]$, the restriction $f|_{[a,x]}$ is continuous on $[a,x]$.

Let $u\in\mathbb{R}$ with $a<u<x$. Then $u$ is an interior point of $[a,x]$, since $a,x\in[a,x]$ and $a<u<x$. Moreover $a<u<x\le b$ shows $u\in(a,b)$, so by hypothesis $f$ is differentiable at $u$ with $f'(u)=0$. By claim 2 of Restriction Stability of Continuity and of the Derivative, applied with the interval $[a,b]$ in the role of $I$ and $[a,x]$ in the role of $J$, the restriction $f|_{[a,x]}$ is differentiable at $u$ with $(f|_{[a,x]})'(u)=f'(u)=0$.

By Mean Value Theorem on a Closed Real Interval, applied to $f|_{[a,x]}$ on $[a,x]$ (note $a<x$), there exists $c$ with $a<c<x$ such that

$$(f|_{[a,x]})'(c)=\frac{f|_{[a,x]}(x)-f|_{[a,x]}(a)}{x-a}=\frac{f(x)-f(a)}{x-a} .$$

The left-hand side is $0$ by the previous paragraph, and multiplying both sides by $x-a\ne0$ gives $f(x)-f(a)=0$, that is, $f(x)=f(a)$. $\blacksquare$