I've been asked to prove that the Sobolev spaces $W^{1,p}(\Omega)$, $\Omega$ open in $\mathbb R^n$, are separable for $1\leq p <\infty$ using the map $$i\colon W^{1,p}(\Omega)\to L^p(\Omega)\times L^p(\Omega)^n \quad u\mapsto (u,\nabla u).$$
I've done something and I would like to know if it is correct.
$W^{1,p}(\Omega)$ is endow with the norm $||u||_{W^{1,p}}:=||u||_{L^p}+||\nabla u||_{{(L^p)}^n}$. Now, I guess that $L^p(\Omega)\times L^p(\Omega)^n$ is endowed with the same norm as $W^{1,p}(\Omega)$ (right?), that makes $L^p(\Omega)\times L^p(\Omega)^n$ a Banach space. With this norm, $i$ is a linear isometry.
Moreover $L^p(\Omega)\times L^p(\Omega)^n$, with the norm above, is a separable metric space. In particular $i(W^{1,p}(\Omega))$ is separable, that is, there exists a countable subset $S\subset L^p(\Omega)\times L^p(\Omega)^n$ which is dense in $i(W^{1,p}(\Omega))$. Since $i$ is an isometry, we have that $i^{-1}(S)$ is a countable dense subset of $W^{1,p}(\Omega).$
Is it correct?