a question in analysis of Sobolev spaces.

Question

I just wanted to know wthether the following is OK or not -
The space $Y=\{u \in H_0^1(\Omega): ||u||_{\infty}\leq M, ||\nabla u||_{2}\leq r\}$ is closed in $X=H_0^1(\Omega)\bigcap L^{\infty}(\Omega)$ because of the following reason. Let $\{u_n\}$ be a Cauchy sequence in $Y$ with $||..||_{1,2}$ norm and hence $||u_n||_{1,2}\leq r$. Hence, $u_n$ converges to $u \in H_0^1(\Omega)$. So, $||u||_{1,2}\leq ||u_n-u||_{1,2}+||u_n||_{1,2}\leq \epsilon+r$ for every $\epsilon > 0$. Hence $||u||_{1,2}\leq r$. Now that $||u_n||_{\infty}\leq M$, hence $0\leq ||u||_{\infty}=\inf_{n}\sup_{x\in \Omega}\{|u_n(x)|\}\leq M$. Thus $Y$ is closed in $X$.