I was reading the following proof from Brezis' book on Sobolev spaces.
Proposition : Let $G\in C^1(\mathbb{R})$ be such that $G(0)=0$ and $|G'(s)|\le M\forall s\in\mathbb{R}$ for some constant $M$. Let $u\in W^{1,p}(\Omega)$ with $1\le p \le \infty$. Then
$$G\circ u \in W^{1,p}(\Omega)\text{ and }\frac{\partial}{\partial x_i}G\circ u = (G'\circ u))\frac{\partial u}{\partial x_i}, i=1,\ldots,N.$$
Proof : We have $|G(s)|\le M|s| \forall s\in\mathbb{R}$ and thus $|G\circ u|\le M|u|$; as a consequence, $G\circ u \in L^p(\Omega)$ and also $(G'\circ u)\frac{\partial u}{\partial x_i}\in L^p(\Omega)$. It remains to verify that
$$\int_{\Omega}(G\circ u)\frac{\partial \phi}{\partial x_i}=-\int_{\Omega}(G'\circ u)\frac{\partial u}{\partial x_i}\phi, \forall \phi\in C^1_c(\Omega).$$
When $1\le p< \infty$, one chooses a sequence $(u_n)$ in $C^{\infty}_c(\mathbb{R}^N)$ such that $u_n\to u$ in $L^p(\Omega)$ and a.e. on $\Omega, \nabla u_n \to \nabla u$ in $L^p(\omega)^N \forall \omega\subset \subset \Omega$. We have
$$\int_{\Omega}(G\circ u_n)\frac{\partial \phi}{\partial x_i} = -\int_{\Omega}(G'\circ u_n)\frac{\partial u_n}{\partial x_i}\phi, \forall \phi\in C^1_c(\Omega).(6)$$ But $G\circ u_n\to G\circ u in L^p(\Omega)$ and $(G'\circ u_n)\frac{\partial u_n}{\partial x_i}\to (G'\circ u)\frac{\partial u}{\partial x_i}$ in $L^p(\omega)\forall \omega\subset\subset \Omega$ (by dominated convergence), so that (6) follows from the above.
I don't really understand why we need $\omega\subset\subset \Omega$ (meaning $\omega$ is compactly contained in $\Omega$). Since $|G'\circ u_n|\le M$ and $u_n, \phi$ are smooth functions with compact support we have the result even if $\omega$ was not compactly contained in $\Omega$ no?
One wants to approximate $W^{1,p}$ functions by smooth functions. For arbitrary $\Omega$ it is not possible to approximate $u$ by functions from $C^\infty_c(\mathbb R^d)$ or $C^\infty_c(\Omega)$. What is possible however, is to approximate $u$ on compact subsets of $\Omega$. This can be proven by simple mollification and use of cut-off function.
In the proof in your post, one simply does not have the existence of a sequence of smooth functions $(u_n)$ such that $\nabla u_n\to \nabla u$ in $L^p(\Omega)$.