Given a Lipschitz domain $\Omega$ in $\mathbb{R}^n$, I know that the space of infinitely differentiable functions with compact support $C^\infty_0 (\Omega) $ is dense in the Sobolev space $W^{1,p}_0(\Omega) = \{v\in W^{1,p}(\Omega): v|_{\partial\Omega} = 0\}$.
What I am wondering is, if I let $ V =\{v\in W^{1,p}(\Omega) : \int_\Omega v dx = 0\}$ and let $T = \{v\in C^\infty(\overline \Omega): \int_\Omega v dx = 0\}$, then is $T$ dense in $V$?
The answer is yes. Let's prove it in this way.
Take $V$ as you defined, first notice that it is any closed subset of $W^{1,p}$ with respect to $W^{1,p}$ norm.
Now, since $\Omega$ process a good boundary, we could find $(u_n)\subset C^\infty(\bar \Omega)$ such that $u_n\to u\in V$ in $W^{1,p}$ norm.
Now we define $$ v_n\equiv u_n-\frac{1}{|\Omega|}\int_\Omega u_n$$ and hence $v_n\in T$.
We have $\nabla v_n=\nabla u_n$ and hence $\nabla v_n\to \nabla u$ in $L^p$. Moreover, $$ \|v_n-u\|_{L^p}\leq \|u_n-u\|_{L^p}+\|\int_\Omega u_n\|_{L^p}\to 0 $$ where $$ \|\int_\Omega u_n\|_{L^p}\to 0 $$ bt LDCT.
Therefore, $T$ is dense in $V$.
As be pointed out in comment, we could use Poincare to prove that $$\|v_n-u\|_{L^p}^p\leq C\|\nabla u_n-\nabla u\|_{L^p}^p\to 0$$