Just wanted to see if my proof of the following is valid:
Let $\{K_i\}_{i=1}^{\infty}$ be compact sets (in some metric space), and let $V$ be an open set such that $$ \bigcap_{i=1}^{\infty} K_i \subset V.$$ Then there exists $m$ such that $$\bigcap_{i=1}^{m} K_i \subset V.$$
Proof: Suppose not. Then for each $n$, there exists $$x_n \in \bigcap_{i=1}^{n} K_i \cap V^c.$$ Let $\{x_n\}_{n=1}^{\infty}$ be the sequence so formed. In particular, this is a sequence in $K_1$ and thus has a convergent subsequence with limit $\hat{x} \in K_1$. Relabel this convergent subsequence as $\{x_n\}_{n=1}^{\infty}$. Now, there exists $k(j)$ so that $\{x_n\}_{n=k(j)}^{\infty} \subset K_j$, and again has a convergent subsequence that converges to $\hat{x} \in K_j$. Since this holds for any $j$, $$\hat{x} \in \bigcap_{i=1}^{\infty} K_i.$$ Since $V^c$ is closed, $\hat{x} \in V^c$, a contradiction.
And maybe a hint as to how to proceed for a purely topological proof?
It’s basically correct. There’s a typo at the very beginning, where you meant to write ‘For each $n$ there exists’ (instead of $j$). And you need to pass to a convergent subsequence only once: the tail sequence $\langle x_n:n\ge k(j)\rangle$ already converges to $\hat x$, since it’s a subsequence of a sequence converging to $\hat x$.
For a proof in case $X$ is not necessarily metric, let $F_n=(K_1\cap K_n)\setminus V$ for each $n\in\Bbb Z^+$, and work in the compact subspace $F_1$. Show that if the conclusion of the theorem fails,
Notice that this argument can be adapted to any family $\mathscr{K}$ of compact sets with non-empty intersection; it doesn’t use the countability of the family.
Added: As Rob Arthan in effect notes in the comments, you do need to assume that $X$ is a $KC$-space, meaning one in which compact sets are closed, in order to ensure that the intersection of a centred family of compact sets is non-empty; this property lies strictly between $T_1$ and $T_2$.