Recovering classical solution from weak one for the Laplace equation Assume that $u:\overline{\Omega} \rightarrow \mathbb{R}$ where $\Omega \subset \mathbb{R}^n$ limited. One can write the Laplace equation on the weak form as 
$$
\begin{equation}
\begin{cases} 
    \int_\Omega \nabla u\nabla \phi =0, &  \forall \phi \in C^1_c(\Omega), \\
    u = 0, &  x\in \partial\Omega .
\end{cases}
\end{equation}
$$
Using Riez's representation theorem and Lax-Milgram one can show that $\exists !u \in H^1_0(\Omega)$ s.t the problem above is satisfied. 
As $C^\infty_c(\mathbb{R}^n)$ is dense in $H^1_0(\Omega)$ and $u=0$ for $x\in \partial\Omega$ can I assume $u \in C^2(\overline{\Omega})$ and therefore a classical solution once that $\Delta u = 0$ a.e?
 A: No, this has to be shown.
There exist solutions to other equations that belong to $H_0^1(\Omega)$, but they are not necessarily twice differentiable. For the Laplacian this is not the case (at least in the interior of the domain), but this has to be proven.
Moreover, you have to distinguish between the cases where you are looking at differentiability in the interior of the domain, or on the boundary. For the Laplacian, the solution will be infinitely many times differentiable in the interior, but not necessarily up to the boundary; it might even not be continuous up to the boundary!
All of this falls under the general title "regularity theory for PDEs", and you can find this in Evans, for example.
Edit: In this particular problem that you are considering, we can show that $u\equiv 0$: to do this, note that since $u\in H_0^1(\Omega)$, there exists a sequence $\phi_n\in C_c^1(\Omega)$ such that $\phi_n\to u$ in $H_0^1(\Omega)$. Then, $$\int_{\Omega}|\nabla u|^2=\lim_{n\to\infty}\int_{\Omega}\nabla u\nabla \phi_n=0,$$ so $u$ is constant in every connected component of $\Omega$. Since $u\in H_0^1(\Omega)$, this implies that $u\equiv 0$.
In general, though, there are many steps involved in order to show that a weak solution as above has good regularity properties.
