FALSE: a monotone function has at most countably many points of non-differentiability

Let $c$ be the Cantor function and fix $x$ in the Cantor set. As in the Cantor-function example, let $u_n\le x\le v_n$ be the stage-$n$ Cantor interval determined by the first $n$ ternary digits of $x$. Then $v_n-u_n=3^{-n}$ and $c(v_n)-c(u_n)=2^{-n}$ by The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$ and The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set. At least one of the two numerator differences $c(v_n)-c(x)$ and $c(x)-c(u_n)$ is at least $2^{-n-1}$. For that choice the corresponding denominator is positive and at most $3^{-n}$, so one of the two secant slopes is at least $\frac{1}{2}(3\mathrm{/}2{)}^n$. Hence $c$ has no finite derivative at $x$. Since the Cantor set is uncountable by The Cantor set is an uncountable subset of $\mathbb{R}$ of Lebesgue measure zero, the nondifferentiability set of $c$ is uncountable.

Therefore a monotone function can have uncountably many nondifferentiability points, so the statement is false.