Let \(X\) be a metric space and \(A\subseteq X\) be closed and \(x_n\to x\) a convergent sequence with \(x_n\in A\) for every \(n\in \mathbb{N}\). Then \(x\in A\).
Proof
Assume \(x\in A^c\). But since \(A^c\) is open we get a contradiction.