Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$
Note that $\psi$ is increasing and bounded (and even continuous).
Consider an enumeration of the rationals $q_n$ and the function
$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$
$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.
However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$
As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Hölder for any $\alpha$.
Edit, by replacing $\log$ by a continuous increasing function that diverges arbitrarily slowly you can construct a continuous, increasing and bounded function with arbitrarily low local regularity.