Suppose that $X$ is a first-countable compact topological space. Consider an infinite sequence $x_1,x_2,\dots,x_n,\dots$ of points in $X.$
**Claim.** There is a point $x\in X$ such that, for every neighborhood $U$ of $x,$ we have $x_n\in U$ for infinitely many $n.$
**Proof of claim.** Assume the contrary. Each point $x$ is covered by an open set $U$ such that $x_n\
otin U$ for all sufficiently large $n.$ By compactness, the space $X$ is covered by finitely many such sets. It follows that $x_n\
otin X$ for sufficiently large $n,$ which is absurd.
Choose a point $x$ as claimed above. Let $V_1\supseteq V_2\supseteq\cdots\supseteq V_n\supseteq\cdots$ be a countable decreasing neighborhood base for $x.$ Choose $n_1\lt n_2\lt n_3\lt\cdots$ with $x_{n_k}\in V_k.$ The subsequence $x_{n_1},x_{n_2},\dots$ converges to $x.$