Let $\Omega\subset \mathbb{R}^n$ be open and bounded. Is it true that $C^{0,1}(\Omega) \subset W^{1,\infty}(\Omega)$? Let $\Omega\subset \mathbb{R}^n$ be open and bounded. If $\partial \Omega \subset C^1$, the result follows from Theorem 4 on page 279 of "Partial Differential Equations" by Evans. Without any assumptions on the boundary, this result was mentioned in a couple of other mse-posts, but I wasn't able to find a reference. So is $C^{0,1}(\Omega) \subset W^{1,\infty}(\Omega)$ true if we don't require smoothness of the boundary of $\Omega$ and if yes, can we argue in the following way?
We have $C^{0,1}(\mathbb{R}^n)=W^{1,\infty}(\mathbb{R}^n)$. By the Kirszbraun theorem, we can extend every function $u\in C^{0,1}(\Omega)$ to $U\in C^{0,1}(\mathbb{R}^n)$ with $u=U$ on $\Omega$. It follows that $U\in W^{1,\infty}(\mathbb{R}^n)$ and thus $u\in W^{1,\infty}(\Omega)$.
 A: The OP's argument is correct, but the extension $u \mapsto U$ is slightly redundant.  As I will show below, the argument (or at least one of the arguments) that shows $C^{0,1}(\mathbb{R}^{d}) \subseteq W^{1,\infty}(\mathbb{R}^{d})$ also allows us to get $C^{0,1}(\overline{\Omega}) \subseteq W^{1,\infty}(\Omega)$ without additional effort.
Suppose $u : \overline{\Omega} \to \mathbb{R}$ satisifes $|u(x) - u(y)| \leq M\|x - y\|$ for all $x,y \in \overline{\Omega}$.  In other words, $u \in C^{0,1}(\overline{\Omega})$.  I claim that $u \in W^{1,\infty}(\Omega)$.
Indeed, if $\varphi \in C^{\infty}_{c}(\Omega)$ and $e \in S^{d-1}$, then
\begin{equation*}
\left| \int_{\Omega} u(y) D\varphi(y) \cdot e \, dy \right| = \lim_{\epsilon \to 0^{+}} \left| \int_{\Omega} \left(\frac{u(x + \epsilon e) - u(x)}{\epsilon} \right) \varphi(x) \, dx \right| \leq M \int_{\Omega} |\varphi(x)| \, dx.
\end{equation*}
Therefore, by $L^{1}(\Omega)$-$L^{\infty}(\Omega)$ duality, for each $e \in S^{d-1}$, there is a $v_{e} \in L^{\infty}(\Omega)$ with $\|v_{e}\|_{L^{\infty}(\Omega)} \leq M$ such that
\begin{equation*}
\int_{\Omega} v_{e}(y) \varphi(y) \, dy = \int_{\Omega} u(y) D\varphi(y) \cdot e \, dy.
\end{equation*}
Using $e = e_{i}$ (a coordinate vector), this shows $u \in W^{1,\infty}(\Omega)$.  (This works if $\Omega$ is any open set, including $\Omega = \mathbb{R}^{d}$.)
Since $\|v_{e}\|_{L^{\infty}(\Omega)} \leq M$ for all $e$, one can go further and show $\|Du\|_{L^{\infty}(\Omega)} \leq M$ (which also follows from the OP's Kirszbraun argument).  The idea is $v_{e} = Du \cdot e$ almost everywhere for each given $e \in S^{d-1}$.  Take $(\tilde{e}_{n})_{n \in \mathbb{N}}$ a dense sequence in $S^{d-1}$ to find $\|Du \cdot e_{n}\|_{L^{\infty}(\Omega)} \leq M$ independent of $n$, hence $\|Du(x)\| \leq M$ almost everywhere in $\Omega$.
