First question. Let $\Omega$ be an open set and $u:\Omega\longrightarrow\mathbb{R}$ be a Lipschitz function in $\Omega$ (in short $u\in\operatorname{Lip}(\Omega)$). I do not understand why $u\in\operatorname{Lip}(\overline{\Omega})$.
Second question. Is it true that if $u\in\operatorname{Lip}(\Omega)$ (and hence by Rademacher Theorem $u$ is differentiable a.e. in $\Omega$), then $\|Du\|_{L^{\infty}(\Omega)}<\infty$?
Thank You
On
The first question doesn't make much sense by itself. If $u$ is only defined on $\Omega$, then you just don't know what its values are on $\overline{\Omega}\setminus \Omega$. Perhaps your question is: Can $u$ be extended to a Lipschitz function on the closure? Let's see:
Take $x_j \to x$, where $x$ is on the boundary of $\Omega$. Then the Lipschitz property means $\{u(x_j)\}_{j=1}^{\infty}$ is Cauchy because $$ |u(x_j) - u(x_i)| \leq L|x_i - x_j| \leq 2L \max\{|x_i- x|, |x_j - x|\}, $$ say, which clearly can be made small by choosing $\min\{i,j\}$ large enough. OK so you have a limit; call it $u(x)$. Now you should also verify separately that this $u$ is Lipschitz.
Second question answer is yes.
First question: do you know that every Lipschitz function $f$ defined on $\Omega$ has a Lipschitz extension to the whole space? This is true if $\Omega$ is a subset of an arbitrary metric space. You can define the extension $\bar f$ by $$\bar f (x) = \inf_{y \in \Omega} ( f(y) + K d(x,y))$$ where $K$ is the Lipschitz constant of $f$ on $\Omega$. By continuity, any extension of $f$ is uniquely defined on $\overline{\Omega}$. Thus $f \in \mathrm{Lip}(\Omega)$ implies $f \in \mathrm{Lip} (\overline \Omega)$.
Second question: wherever a Lipschitz function is differentiable each partial derivative is bounded by the Lipschitz constant, so the gradient is essentially bounded.