An elementary non-existence proof may be of interest.
Let $f:[a,b]\to\mathbb{R}$ an increasing, continuous, and not absolutely continuous function: I claim there exists a point $c\in[a,b]$ where the Dini derivative $D^*f(c):=\limsup_{x\to c}\frac{f(x)-f(c)}{x-c}$ is infinite. For the proof, we may assume w.l.o.g. that $f$ is strictly increasing (we may just consider $x+f(x)$).
By definition, since $f$ is not absolutely continuous, there exists a sequence of sets $J_n\subset [a,b]$ such that each $J_n$ is a finite union of intervals, $|J_n|\to0$, and $|f(J_n)|$ is bounded away from $0$.
Let $J^*_n\subset J_n$ be the union of all components $I$ of $J_n$ such that $|f(I)|\ge\frac12\frac {|f(J_n)|}{|J_n|}|I|$. Clearly, $|f(J_n\setminus J_n^*)|\le\frac {|f(J_n)|}2$, so the sets $f(J_n^*)$ have length $|f( J_n^*)|\ge\frac {|f(J_n)|}2$ bounded away from $0$ too. This implies ($\dagger$) that some $p \in f([a,b])$ belongs to infinitely many $f( J_n^*)$. Therefore $c:=f^{-1}(p)$ belongs to infinitely many $J_n^*$; for these indices, let $[\alpha_n,\beta_n]$ be the component of $c$ in $J_n$. So by construction both $\beta_n$ and $\alpha_n$ converge to $c$ along a subsequence; and w.r.to this subsequence, $\max\big\{\frac{f(\beta_n)-f(c)}{\beta_n-c},\frac{f(c)-f(\alpha_n)}{c-\alpha_n}\big\}\ge\frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}=\frac{|f(I_n)|}{|I_n|}\to+\infty$, proving that $D^*f(c)=+\infty$.
($\dagger$) It's the well-known argument from measure theory: For a finite measure space $(X,\mathcal{S},\mu)$, if for $n\in\mathbb N$ one has $Y_n\in\mathcal{S}$ and $\mu(Y_n)\ge \lambda$, then $\overline{Y}:=\limsup_{n\to\infty}Y_n:=\cap_{k\in\mathbb N}\cup_{n\ge k}Y_n$ also has $\mu(\overline{Y})\ge \lambda$. Here $Y_n:=f( J_n^*)$.