Compact sets in Hausdorff spaces

The proof fixes \(p\notin K\). For each \(x\in K\), Hausdorff gives disjoint open sets \(U_x\ni p\) and \(V_x\ni x\). The \(V_x\) cover \(K\); by compactness \(V_{x_1},\dots,V_{x_n}\) suffice, and \(U=\bigcap U_{x_i}\) is an open set containing \(p\) that misses \(K\). Thus \(X\setminus K\) is open.

The Hausdorff hypothesis is indispensable. In \(\{a,b,c\}\) with open sets \(\varnothing,\{a\},X\), the set \(\{a\}\) is finite, hence compact, but not closed; and in the cofinite topology on \(\mathbb{N}\) the evens are compact and not closed.

The other half says a closed subset \(F\) of a compact space \(X\) is compact: an open cover of \(F\) is completed with \(X\setminus F\) to cover \(X\), a finite subcover is extracted, and \(X\setminus F\) is discarded. This half needs no Hausdorff. Together they give that in a compact Hausdorff space, compact and closed mean the same.

Immediate consequences: the intersection of compact sets in a Hausdorff space is compact, and in \(\mathbb{R}\) every compact set is closed, which is the easy half of Heine-Borel.