Assume that $\Omega\subset\mathbb{R}^{n}$ is a bounded domain and that $u\in W_{0}^{1,p}(\Omega)\cap C(\Omega)$ . Let $N\not=\emptyset$ be a nodal domain of $u$ , that is, a connected component of the set $\left\{ x\in\Omega:u(x)>0\right\}.$ Will the restriction of $u$ to $N$ belong in $W_{0}^{1,p}(N)$ ?
If we in addition assume that $u\in C(\overline{\Omega})$ and $u\equiv0$ on $\partial\Omega$, then the above statement should follow somewhat easily, but can this condition be left out or relaxed?
Edit: I might have found a solution. Here is a sketch of the proof I thought of but I am not so sure.
Proof. We first define a sequence of functions $\psi_{n}$ by setting $\psi_{n}:=\max(u^{+}-\frac{1}{n},0),$ where $u^{+}$ denotes the positive part of $u$. Then $\psi_{n}\in W^{1,p}(\Omega)$ and $\psi_{n}\rightarrow u^{+}$ in $W^{1,p}(\Omega)$. By continuity of $u$ and the definition of the set $N$ , there exists an open set $A_{n}$ such that $$\psi_{n}\equiv0\,\mbox{in }A_{n}\,\mbox{and }\partial N\cap\Omega\subset A_{n}.$$ Since $u^{+}\in W_{0}^{1,p}(\Omega)$ is nonnegative, there exists a sequence of nonnegative functions $\varphi_{n}\in C_{c}^{\infty}(\Omega)$ such that $\varphi_{n}\rightarrow u^{+}$ in $W^{1,p}(\Omega)$ . For each of these there exists an open set $B_{n}$ such that $$\varphi_{n}\equiv0\,\mbox{in }B_{n}\,\mbox{and }\partial\Omega\subset B_{n}.$$ We now define $$u_{n}:=\min\left(\psi_{n},\varphi_{n}\right).$$ Then at least $u_{n}\in W^{1,p}(\Omega)$ and also $u_{n}\rightarrow u^{+}$ in $W^{1,p}(\Omega)$. Since $\psi_{n}$ and $\varphi_{n}$ are both nonnegative, it holds that $u_{n}\equiv0$ in $A_{n}\cup B_{n}\supset\left(\Omega\cap\partial N\right)\cup\partial\Omega\supset\partial N$. The function $u_{n|E}$ is thus compactly supported in $N$ and therefore belongs to $W_{0}^{1,p}(N)$.
To prove that $u_{|N}\in W^{1,p}_0(N)$, it is sufficient to prove that $u^+_{|N}\in W_0^{1,p}(N)$. Your proof can be divided in three steps:
I - Consider the sequence $\psi_n=\max\{u^+-1/n,0\}$. Once $u$ is continuous, for each point $p\in \partial N\cap \Omega$, there is a neighbourhood $U_{n,p}$ such that $\psi_n(x)=0$ for $x\in U_{n,p}$, however, this does not need to be true for $p\in \partial\Omega$, so we need the next step.
II - Once $u^+\in W_0^{1,p}(\Omega)$, there is a sequence $\varphi_n\in C_0^\infty(\Omega)$ converging to $u^+$ in the $W^{1,p}(\Omega)$ norm. Note now that for all $p\in \partial\Omega$, there is a neighbourhood $U_{n,p}$ such that $\varphi_n(x)=0$ for $x\in U_{n,p}$, however, this does not need to be true for $p\in \partial N\cap \Omega$, so we need the next step.
III - Combine steps I and II, in the form $u_n=\min\{\psi_n,\varphi_n\}$, to conclude that $u_n$ has compact support in $N$ and $u_n\to u^+$ in $W^{1,p}(\Omega)$, therefore, the claim is true.
In the end, your proof seem to be correct. It just need some minor modifications.