Does a Lipschitz continuous function have one-sided derivatives everywhere? Let $f\colon\mathbb R\to\mathbb R$ satisfy the Lipschitz condition: there exists $K\geq 0$ such that for all $x,y\in\mathbb R$, we have $|f(x)-f(y)|\leq K\cdot |x-y|$. Is it true that $f$ has one-sided derivatives everywhere? I.e., that the limits
$$\lim_{h\nearrow 0}\frac{f(x+h)-f(x)}{h}\quad\text{and}\quad\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h}$$
exist for every $x$?
I know that both exist a.e. since the Lipschitz condition implies that $f$ is differentiable a.e., but I would like to know if it is not only true a.e., but everywhere. After working with this for some time, I got the feeling that it might be.

No. Counterexample: 
Let $g : [1,2] \to \mathbb R$ be the graph which linearly connects the points 
$$(1, 1), (1.5, -1.5), (2, 2).$$
Define $f [0,1] \to \mathbb R$ be 
$$f(x) = \begin{cases} 0 & \text{if } x=0,\\ \frac{1}{2^{n+1}}\  g(2^{n+1} x) &\text{if }x\in [2^{-(n+1)}, 2^{-n}), n =0, 1, 2\cdots.\\
\end{cases}$$
Then $f$ is Lipschitz since it has bounded derivatives (except at some countable points where the derivatives does not exists). However, 
$$\lim_{h\to 0^+} \frac{f(h) - f(0)}{h}$$
does not exist since 
$$ \frac{f(2^{-(n+1)}) - f(0)}{2^{-(n+1)}}  =1, \ \ \ \frac{f(2^{-(n+1)} 1.5) - f(0)}{2^{-(n+1)} 1.5} = -1$$
