One can easily show that $W_0^{1,2}((a,b)) \subset C^0((a,b))$ for any finite interval $(a,b)$. Intuitively $W_0^{1,2}((a,b))$ should contain more functions than $C_0^0((a,b))$, but how to prove that?
I guess a function which is nowhere differentiable in $(a',b') \subset (a,b)$ should do the trick, but I have, sadly, no idea how to prove that.
Any hints?
Let $f$ be the cantor function on $[0,1]$. Let $g(x) = (1-x) f(x)$ on $[0,1]$. Then, $g$ is continuous and $g(0) = g(1) = 0$.
For almost every $x\in[0,1]$ we have $$ g'(x) = \underbrace{(1-x) f'(x)}_{=0} - f(x) = -f(x) < 0.$$ If $g$ is (locally) absolutely continuous, then $g$ would be strictly decreasing, which contradicts $g(0) = g(1)$. Thus, $g$ is not locally absolutely continuous and therefore not in $W^{1,2}[0,1]$.
Now, extend $g$ with 0 on $(-\epsilon, 1+\epsilon)$ for $\epsilon > 0$ if you like.