Let $A_0 = \{w \in C^2(\bar U) | w = 0 \text{ on } \partial U \}$ and define $$ I(w)=\int_U \frac12 |\nabla w|^2 -wf. $$
We can show that $I$ is bounded from below on $A_0$. Hence there exists a sequence $(u_n) \subset A_0$ with $I(u_n) \longrightarrow \inf_{w \in A_0} I(w)$.
The script now states that using the parallelogram identity, we can show that $(\partial_{x_i} u_n)$ is cauchy for every $i$. I don't see how this works, can someone explain please?
The idea is that if the difference $\nabla u_n-\nabla u_m$ has large norm, then the average $(u_n+u_m)/2$ has much smaller $I$ than either $u_n$ or $u_m$. This is a consequence of strong convexity of the functional, which indeed has to do with the parallelogram identity.
Let $u=(u_n+u_m)/2$. From $$ |\nabla u_n+\nabla u_m|^2 + |\nabla u_n-\nabla u_m|^2 = 2(|\nabla u_n|^2 + |\nabla u_m|^2) $$ we get $$ |\nabla u|^2 = \frac12(|\nabla u_n|^2 + |\nabla u_m|^2) - \frac14 |\nabla u_n-\nabla u_m|^2 $$ Integration yields $$ I(u) = \frac12(I(u_n) + I(u_m)) - \frac18\int_U |\nabla u_n-\nabla u_m|^2 $$ Rearrange this as $$ \frac18\int_U |\nabla u_n-\nabla u_m|^2 = \frac12(I(u_n) + I(u_m)) - I(u) \le \frac12(I(u_n) + I(u_m)) - \inf_{w \in A_0} I(w) \to 0 $$ and the result follows.