Proposition 4

Every compact subset of a Hausdorff space is closed.

Let $A$ be a compact subset of a Hausdorff space $X$, and let $x$ be any point of $\overline{A}$; we have to show that $x \in A$. By hypothesis, every neighbourhood of $x$ meets $A$, and therefore the neighbourhood filter $\mathfrak{B}$ of $x$ in $X$ induces a filter $\mathfrak{B}_A$ on $A$; $A$ is compact whence $\mathfrak{B}_A$ has a cluster point $y \in A$. Since the filter $\mathfrak{B}$ is coarser than the filter on $X$ generated by $\mathfrak{B}_A$ (considered as a filter base on $X$), $y$ is also a cluster point of $\mathfrak{B}$; hence $y = x$, because $\mathfrak{B}$ converges to $x$ in $X$ and $X$ is Hausdorff (§ 8, no. 1, Proposition 1).