Is $\nabla u \in L^{\infty}$ if $u$ is bounded $C^{0}$? I would like to prove something of the form $|A_{1}(u)| \leq c \lVert u \rVert_{L^{\infty}}$ and $|A_{2}(u)| \leq c \lVert \nabla u \rVert_{L^{\infty}}$ for some operators $A_{1},\ A_{2}$ and arbitrary constant $c$. I am working on domain $\Omega \subset \mathbb{R}^{2}$ which is bounded and open, and has $C^{1}$ boundary. I know that $u \in W^{1,p}(\Omega)$ for $p > 2$. From this I can use general Sobolev embeddings to show that $W^{1,p} \hookrightarrow C^{0,\gamma}$ for some $\gamma < 1$. If I now also assume that $u$ is a bounded function on $\Omega$ such that $u \in L^{\infty}(\Omega)$ I would like to show that $\nabla u \in L^{\infty}(\Omega)$, which seems as though it should also be valid. I only have intuition on this, but it seems as though a continuous function should have a bounded derivative a.e. if it is bounded itself.
I am not sure how to show the above result, though. I think that what I need to do is to show that if the Cauchy sequence $\{u_{n}\}$ for $u_{n} \in L^{\infty}(\Omega) \cap C^{0,\gamma}(\Omega)$ converges to $u \in L^{\infty} \cap C^{0,\gamma}(\Omega)$ (i.e. $u_{n} \rightarrow u$ in $L^{\infty}$) then this also implies that $\{\nabla u_{n}\}$ is a Cauchy sequence with $\nabla u_{n} \rightarrow \nabla u^{*}$ and then show that $u^{*} = u$. However, I am not sure how to do this at all. Any help would be appreciated.

Let $\Omega = \overline{B_1(0)}$ and $u(x) := \sqrt{1-\Vert x\Vert^2}$.
Then $u\in C^0(\Omega) \cap L^\infty(\Omega) \cap C^\infty(\Omega^\circ)$, but
$\nabla u \notin L^\infty(\Omega)$, thus this $u$ serves as a counter-example.
