I am reading Gilbarg and Trudinger's Elliptic PDE of the second order. And I am confused about the proof of Thm. 11.4. 
I don't know how to prove the operator $T$ is continuous. After showing $u = Tv$, I think it only derives that $\{T\bar{v}_m\}$ converge to $u$. But how to show that $\{Tv_m\}$ also converges to $u$?
The argument uses the following fact: $x_{n} \to x$ if and only if every subsequence of $(x_{n})$ has a further subsequence that converges to $x$. For a proof, see this question (the proofs are on $\mathbb{R}$, but if not difficult to adapt them to metric spaces).
Let $v_{m} \to v$. To show that $T$ is continuous, we have to show that $Tv_{m} \to Tv$. Since $\{Tv_{m}\}$ is precompact in that space, this says that every subsequence has a further subsequence that converges to some $u \in C^{2}(\overline{\Omega})$. Then they prove that $u$ solves the PDE, that is, $u=Tv$. Thus, they just have showed that any subsequence of $\{Tv_{m}\}$ have a subsubsequence $\{T\overline{v}_{m}\}$ that converges to $u=Tv$. Therefore, by the given fact, this means that $Tv_{m} \to Tv$, that is, $T$ is continuous.
