The following is the elementary answer that Pietro Majer gave to this question on MO. I copy the answer here so we can consider this question answered.
Consider the nested family of open nbd's of $(0,1)\cap\mathbb{Q}\ :$ $$A_\epsilon:=\cup_{n\in\mathbb{Z} _ + } (r_n- \epsilon 2^{-n/3},r_n+ \epsilon 2^{-n/3})\ , \qquad \epsilon > 0\ . $$
So $|A _\epsilon|=O(\epsilon)$ and $A:=\cap _ {\epsilon > 0} A _ \epsilon$ has measure zero. Let $x \in (0,1) \setminus A$: There exists $\epsilon > 0$ such that for any $n\in\mathbb{Z}_+$ there holds $ \epsilon 2^{-n/3}\le |x-r_n|$. Thus, for any $y\in (0,1)$
$$|f(x)-f(y)|\le \sum_{|x- r _ n|\le|x- y| } 2^{-n}= \frac{1}{\epsilon^2}\sum_{|x- r _ n|\le|x- y| } 2^{-n/3}(\epsilon 2^{-n/3})^2\le $$
$$\le \frac{1}{\epsilon^2}\bigg(\sum_{n=1}^\infty 2^{-n/3}\bigg)|x-y|^2= \frac{|x-y|^2}{\epsilon^2(2^{1/3}-1))}\ ,$$ showing that $f'(x)=0\ .$