You can't find a reference because it's false. Rademacher's theorem (Lebesgue-almost everywhere differentiability) is the best one can do.

In fact, *for every Lebesgue-null set $E \subset \mathbf{R}$, you can construct a Lipschitz function $f: \mathbf{R} \to \mathbf{R}$ that is not differentiable at any point of $E$.* [ACP10]

The reference for this is Theorem 1.1 in... the same paper you linked in your question. The result is on page 1. I don't think they give a proof there, but elsewhere, Preiss gives the argument in some [notes](https://homepages.warwick.ac.uk/~masfay/talks/helsinki_07_08_slides.pdf) from a talk given in Helsinki. The proof basically goes as follows.

*Proof.*
Construct some nested sequence of open sets $\mathbf{R} = G_0 \supset G_1 \supset \cdots \supset E$ that are rapidly shrinking: for every connected component $C$ of $G_k$, the next open set has
\begin{equation}
\lvert G_{k+1} \cap C \rvert \leq 2^{-k-1} \lvert C \rvert.
\end{equation}
(Let us say additionally that $\lvert G_1 \rvert \leq 1$.) You can find such a sequence because the Lebesgue measure is outer regular.

The sets $(G_k \setminus G_{k+1} \mid k \in \mathbf{N})$, together with $E$, partition the real line, and we define the function
$\psi: x \mapsto (-1)^k$ if $x \in G_k \setminus G_{k+1}$. This is bounded and measurable. (We need not define $\psi$ on $E$ as it is a null set.)

Then the desired function is $f: x \in \mathbf{R} \to \int_0^x \psi$. This is Lipschitz, but not differentiable at any point of $E$.

Essentially, the reason is the following: 
take an arbitrary point $x \in E$, and let, for each $k \in \mathbf{N}$, $(a_k,b_k)$ be the connected component of $G_k$ containing $x$. Then the derivative of $f$ on $(a_k,b_k)$ is equal to $(-1)^k$ on a large portion of the interval. Certainly this is the case on $(a_k,b_k) \setminus G_{k+1}$, and therefore
\begin{equation}
\lvert f(b_k) - f(a_k) - (-1)^k (b_k - a_k) \rvert
\leq 2 \lvert (a_k,b_k) \setminus G_{k+1} \rvert
\leq 2^{-k} (b_k - a_k).
\end{equation}

If you now repeat the same calculation for the next term, you  find that
\begin{equation}
\lvert f(b_{k+1}) - f(a_{k+1}) - (-1)^{k+1} (b_{k+1} - a_{k+1}) \rvert \leq 2^{-k-1} (b_{k+1} - a_{k+1}). 
\end{equation}
If you divide this equation through by $b_{k+1} - a_{k+1}$, respectively the previous equation through by $b_k - a_k$, you will see that the two combined are incompatible with differentiability of $f$ at $x$. Q.E.D.





[ACP10] G. Alberti, M. Csörnyei, and D. Preiss. *Differentiability of lipschitz functions, structure of null sets, and other problems.* In Proceedings of the ICM 2010, pages 1379-1394.