Concavity near the boundary of the distance function I was reading the paper

Quelques remarques sur les problemes elliptiques quasilineaires du
second ordre, P. L. Lions, Journal d’Analyse Mathématique volume 45,
pages 234–254(1985)

and on page 251 he claims that (using it in the proof), the distance function near the boundary is concave for a general open bounded domain, which I don't think it is true. I can get this fact if the domain is convex, is there any result saying this for a non-convex domain? Say, interior ball condition is enough?

Edit: On page 251, here is the part I suspected that he used that fact:
\begin{align}
|u(\overline{x}-s\mathbf{n}(\overline{x}))-u(y)| &= |u(\overline{x}-s\mathbf{n}(\overline{x}))-u(\overline{y} - s\mathbf{n}(\overline{y}))|\\
&\leq \frac{C}{s^{1-\theta}}|\overline{x} - s\mathbf{n}(\overline{x}) - y|.
\end{align}
My reasoning is that, he has $|\nabla u(x)\leq \frac{C}{d(x)^{1-\theta}}$ earlier. Therefore to connect the line between $\overline{x}-s\mathbf{n}(\overline{x})$ and $y = \overline{y}-s\mathbf{n}(\overline{y})$ we need that every point on this line segment need to have distance bigger or equal to the distance from $\overline{x}-s\mathbf{n}(\overline{x})$ or $y$, in short, we want
\begin{equation}
d_{\partial \Omega}\Big(\lambda\big(\overline{x}-s\mathbf{n}(\overline{x})\big)+(1-\lambda )y\Big) \geq s
\end{equation}
provided that at $\lambda = 0$ or $\lambda  = 1$ we have
\begin{equation}
d_{\partial\Omega}(y) = d_{\partial\Omega}(\overline{x}-s\mathbf{n}(\overline{x})) = s.
\end{equation}
We could say quasi-concave is enough, but I do believe he meant concave.
 A: The domain is assumed regular therefore at least Lipschitz in my opinion, and this is sufficient to ensure quasi-convexity of the domain. This means that for $z_1,z_2\in \Omega$ you have $\mathrm{d}_\Omega(z_1,z_2) \lesssim_\Omega |z_1-z_2|$ where $\mathrm{d}_\Omega$ is the geodesic distance (minimal length of a continuous $\Omega$-valued path between the two points).
Here you have the assumption $s\geq |x-y|$ so WLOG $s>0$. In particular $\overline{x}-s \mathbf{\overline{x}}$ and $y$ are two points of $\Omega$ and you can therefore use your estimate on the gradient on the given path.
For a reference about quasi-convexity you can check Sections 2.5.1 and 2.5.2 of

Brudnyi, A.; Brudnyi, Y., Methods of geometric analysis in extension and trace problems.
Volume 1. Monographs in Mathematics, 102. Birkh¨auser/Springer, 2012.

EDIT : sorry, I got confused in the previous answer. You invoke the quasi-convexity of $\partial\Omega$ to get the existence of a map $\gamma:[0,1]\rightarrow\partial\Omega$ linking $\overline{x}$ to $\overline{y}$. Then, $\gamma_s(\sigma):=\gamma(\sigma)-s\mathbf{n}(\gamma(\sigma))$ maps $\overline{x}-s\mathbf{n}(\overline{x})$ to $y$ and satisfies $\mathrm{d}_{\partial_\Omega}(\gamma_s)\geq s$ all the way. The length of $\gamma_s$ should be (I confess, I did not check it) comparable to the one of $\gamma$, which is, by quasi-convexity, comparable to $|\overline{x}-\overline{y}|$, itself comparable to $|x-y|$ (beginning of page 251).
That's just the idea that in $\Omega= \mathbf{R}^2\setminus \text{B}(0,1)$ to connect two points $x$ and $y$ of norm $1+\varepsilon$, you do not follow the straightline but a portion of arc (of radius $1+\varepsilon$).
