Lipschitz continuity implies differentiability almost everywhere.

I am running into some troubles with Lipschitz continuous functions.

Suppose I have some one-dimensional Lipschitz continuous function $f : \mathbb{R} \to \mathbb{R}$. How do I prove that its derivative exists almost everywhere, with respect to the Lebesgue measure?

I found on other places on the internet that any Lipschitz continuous function is absolutely continuous, and that this directly implies that the functions is differentiable almost everywhere. I don't quite see how this argument goes, though.

Any help with giving such a proof, or redirecting me to a source where I can find one, would be greatly appreciated.

An excellent reference for the properties of Lipschitz functions is Lectures on Lipschitz Analysis by Juha Heinonen. The differentiability a.e. is Theorem 3.2, page 19. I state the steps of the proof here:

1. Lipschitz continuity implies having bounded variation.
2. A function of bounded variation can be written as the difference of two increasing functions
3. An increasing function is differentiable almost everywhere: this is the main step of the proof, which uses the Vitali covering theorem. Concerning this step, see also $f$ continuous, monotone, what do we know about differentiability?