$f(x)=sum_{n=1}^{infty} {4^{1-n}} {h(4^{n-1}x)}$ is continuous and nowhere monotonic,$h(x)=vert x vert$ with period 1,$vert x vert leq 1/2$

Here is a fairly detailed proof.

Proof that f is continuous:

This is the easier half of the proof. First note that $ h(x)leq1$ for all x. Thus, we have $$ Biglvertfrac{h(4^{n-1}x)}{4^{n-1}}Bigrvert leq frac{1}{4^{n-1}}$$.

Since $sum_{n=1}^{infty}frac{1}{4^{n-1}}<infty$, the Weirstrass M-test implies $sum_{n=1}^{N}frac{h(4^{n-1}x)}{4^{n-1}}$ converges uniformly to f as N goes to $infty.$ Thus, f is the uniform limit of continuous functions, so f is continuous.

Proof that f is not monotone on any interval: This is the more delicate part. First, we fix some notations. Let $$f_{1}(x)=h(x)$$ and $$f_{n}(x)=frac{f_{1}(4^{n-1}x)}{4^{n-1}},$$ so that $$ f(x) = sum_{n=1}^{infty}f_{n}(x)=sum_{n=1}^{infty}frac{f_{1}(4^{n-1}x)}{4^{n-1}}.$$ Note that $f_{n}$ is bounded above by $frac{1}{2}4^{-n+1}$ and is periodic with period $4^{-n+1}$.

Now let $$A= left{xinmathbb{R}:x=k 4^{-m}, kinmathbb{Z},m=0,1,2,.. right}$$ and note that $A$ is dense in $mathbb{R}$ by the Archimedean property of $mathbb{R}$ for instance. Now note that for $ain A$, say $a=k 4^{-m}$, we have $$f_{n}(a)= 0$$ for all $n>m$, since $f_{n}$ vanishes at integers. Thus for such $a$, we have $$f(a)= sum_{n=1}^{m}f_{n}(a).$$ Now let $m$ be a positive integer and let $$x_{m}=4^{-2m-1}.$$ Then for similar reasons as above, we have $$f_{n}(a+x_{m})=0$$ for all $n>2m+1.$ It then follows that $$f(a+x_{m})-f(a)=sum_{n=1}^{m}f_{n}(a+x_{m})-f_{n}(a) + sum_{n=m+1}^{2m+1}f_{n}(a+x_{m}) geq -mx_{m}+(m+1)x_{m} = x_{m}>0.$$ Similarly, $$ f(a-x_{m})-f(a) geq -mx_{m}+(m+1)x_{m}=x_{m}>0.$$ Thus by the density of $A$, it follows $f$ is not monotone on any interval.