Show that an increasing function has derivative $0$ a.e.

First note that $F$ is the cdf of the random variable $X:=\sum_1^{\infty} 2^{-n} \xi_n$ where the $\xi_n$ are i.i.d. Bernoulli$(p)$ random variables. Indeed, it is clear that that $X = \frac12\xi_1+\frac12 Y$, where $Y$ has the same distribution as $X$ and is independent of $\xi_1$. This gives the relation $$P(X\le x) = P(X\le x|\xi_1=0)P(\xi_1=0)+P(X \le x|\xi_1=1)P(\xi_1=1) $$$$= (pP(Y\leq 2x)+q\cdot 0)1_{\{x \le 1/2\}} + (p\cdot 1 +qP(Y\leq 2x-1))1_{\{x >1/2\}},$$ which is exactly the relation for $F$.

Now note by the strong law of large numbers that $X$ is supported on the set of real numbers whose binary expansion has asymptotic density $p$ of $1$'s (or equivalently, has asymptotic density $q$ of $0$'s).

But the set of all such real numbers has Lebesgue measure zero. Indeed, if we uniformly sample a real number from $[0,1]$, then its binary digits are i.i.d. Bernoulli$(1/2)$, thus almost surely the asymptotic density of $1$'s is $1/2$, not $p$.

We conclude that the law of $X$ is singular with respect to Lebesgue measure, which is equivalent to the condition that $F'=0$ a.e..