I tried to use using Green's formula to prove $\triangle u=0$. But it turned out to be wrong because $u=0$ doesn't imply $\triangle u=0$ on the boundary.
I am a novice in graduate pde, so I didn't learn hilbert spaces yet. I am wondering is there a proof using only harmonic functions?
The second Green's identity says that $$\int_\Omega f_1\Delta f_2-\int_\Omega f_2\Delta f_1=\int_{\partial\Omega}f_1\frac{\partial f_2}{\partial\nu}-\int_{\partial\Omega}f_2\frac{\partial f_1}{\partial\nu}.$$ Putting $f_1=u$ and $f_2=\Delta u$, we have $$\int_\Omega u\Delta (\Delta u)-\int_\Omega (\Delta u)^2=\int_{\partial\Omega}u\frac{\partial (\Delta u)}{\partial\nu}-\int_{\partial\Omega}\Delta u\frac{\partial u}{\partial\nu}.$$ By assumption that $u=\frac{\partial u}{\partial\nu}=0$ on $\partial\Omega$, the right hand side is zero, and the left hand side is equal to $-\int_\Omega (\Delta u)^2$ since $\Delta (\Delta u)=0$ in $\Omega$ by assumption. This implies that $\Delta u=0$ in $\Omega$.
Now $\Delta u=0$ in $\Omega$ and $u=0$ on $\partial\Omega$, we can conclude that $u=0$ in $\Omega$ using first Green's identity. I think you can do it from here.