Proof . Since $W^{1,p}$ is a Banach space, and $f_i\weakto f$, then $f_i$ is uniformly bounded in $W^{1,p}$ by Banach-Steinhaus Theorem. Since $W^{1,p}$ is compactly embedding into $L^p$, we have, by passing to subsequence, $f_i\to g$ in $L^p$ for some $g\in L^p$.
For any $\phi\in C_0^\infty$, define a $T_\phi$ as
$$T_\phi(f):=\int_\Omega f\phi,\quad\forall f\in W^{1,p},$$
then $T_\phi\in (W^{1,p})’$, i.e., in the dual space of $W^{1,p}$. Thus, by weakly convergence, we have
$$T_\phi(f_i)\to T_\phi(f)\Longleftrightarrow \int_\Omega f_i\phi\to\int_\Omega f\phi.$$
On the other hand, $f_i\to g$ strongly in $L^p$, thus by Holder inequality, we have
$$\int_\Omega f_i\phi\to\int_\Omega g\phi.$$
In conclusion, we have
$$\int_\Omega f\phi=\int_\Omega g\phi,$$
and we finish the argument by the density of $C_0^\infty$ in $L^p$.