Let $f$ be non linear in the interval $[-1,1]$. We can suppose $f(-1)=f(1)=0$ and $f(x)>0$ for some $x\in(-1,1)$. Now let $N$ be so big that the closed ball of center $(0,-N)$ and with radius $\sqrt{N^2+1}$ doesn't contain the whole graph of $f$. Let $(x_0,f(x_0))$ be one of the points of the graph furthest from $(0,-N)$. Then $f$ is concave at $x_0$.
Indeed, the whole graph of $f$ is contained in the ball of center $(0,-N)$ and boundary passing through $(x_0,f(x_0))$. This implies that the derivative of $f$ at $x_0$ has to be the slope of the tangent to that ball at $(x_0,f(x_0))$, and that tangent only intersects the graph of $f$ at $(x_0,f(x_0))$ (in fact it only intersects the ball at that one point).
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 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.
Best Answer
I just stumbled across your question. I have no idea how the proofs of these results go—and I am inclined to believe that they would indeed also prove the version you seek—but here's a way to deduce the local result from that on $\mathbf{R}^d$.
Let $X \subset \mathbf{R}^d$ be an open, convex set, $f: X \to \mathbf{R}$ be a convex function, and $E \subset X$ be the set of points where $f$ is not differentiable.
In general $f$ cannot be extended to a function defined on $\mathbf{R}^d$. However, if it were bounded and Lipschitz, then it could be [Thm. 4.1,Yan12]. Indeed, under these assumptions the extension $\bar{f}: \mathbf{R}^d \to \mathbf{R}$ can be defined by setting \begin{equation} \bar{f}(z) := \operatorname{sup} \{ t f(x) + (1-t) f(y) \mid x,y \in X, t \geq 1, z = t x + (1-t) y \} \end{equation} for all points $z \in \mathbf{R} \setminus X$.
Let $x \in X$, and $r > 0$ be so small that $D_r(x) \subset \subset X$. Write $f_{x,r}: D_r(x) \to \mathbf{R}$ for the restriction of $f$ to this disk. As convex functions are locally Lipschitz, $f_{x,r}$ is Lipschitz and bounded, and we may thus extend it to $\bar{f}_{x,r} :\mathbf{R}^d \to \mathbf{R}$.
The global version gives that the Hausdorff dimension of $E \cap D_r(x)$ is at most $d-1$. The conclusion follows after covering $X$ with a countable collection of such disks.