= Compact subset of a Hausdorff space
Let $K\subseteq X$, where the ambient <topological space> $X$ is <Hausdorff> and $K$ is <compact> in its <subspace topology>. Then $K$ is a <closed set> in $X$: the <complement> $X\setminus K$ is an <open set> in $X$. The ambient space need not be <compact>. For example, $[0,1]$ is a <compact> <closed interval> in the non-<compact> <Hausdorff> <real line>.<Point-Set Topology by Romyar Sharifi/Chapter 3: Connected and compact spaces>[Lemma 3.2.7 and Proposition 3.2.8]{ref}
The ambient <Hausdorff> hypothesis matters even when $K$ itself is <Hausdorff>. In the <Sierpiński space> $S=\{0,1\}$, the <open sets> are $\varnothing,\{1\},S$, so the <closed sets> are $\varnothing,\{0\},S$. Thus $K=\{1\}$ is <compact> and <Hausdorff> in its one-point <subspace topology>, but is not <closed> in $S$: its <complement> $\{0\}$ is not an <open set>. Every <open> <neighborhood> of $0$ contains $1$, so separation from $K$ fails. In contrast, the <discrete topology> on the same underlying <set> is <Hausdorff> and makes both <singletons> <closed>.
\Image[/sierpinski-compact-subset-closedness.png]
{title=A <compact subset> is <closed> in the two-point <discrete space> but need not be <closed> in the <Sierpiński space>}
The diagram lists every <open set> and <closed set>, so the distinction can be checked directly. All <subsets> in both panels are <compact>: for any <open cover> of a <finite set>, choose one covering <open set> per point, obtaining a <finite subcover>. <Compactness> alone therefore cannot distinguish the two panels.
Here is the finite-selection mechanism in the general proof. For $x\in X\setminus K$ and each $k\in K$, the <Hausdorff> property gives disjoint <open> <neighborhoods> $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ give an ambient <open cover> of $K$. Select a <finite subcover> $U_{k_1},\ldots,U_{k_m}$ and set
$$
U=\bigcup_{j=1}^{m}U_{k_j},\qquad V=\bigcap_{j=1}^{m}V_{k_j}.
$$
Then $K\subseteq U$, $x\in V$, and $U\cap V=\varnothing$. The finite <intersection> makes $V$ an <open set>; an infinite <intersection> would not guarantee this. Thus every point of $X\setminus K$ has an <open> <neighborhood> contained in that <complement>, proving that $K$ is <closed>. The <empty set> is already <closed>.
This theorem combines with the <continuous image of a compact space> theorem to make every <continuous map> from a <compact space> to a <Hausdorff space> a <closed map>: a <closed subset> of its domain is <compact>, and its <image> is <compact> and hence <closed> in the target. This is the reason the inverse in the <compact-to-Hausdorff continuous bijection theorem> is <continuous>. The converse implication, from <closed> to <compact>, needs additional hypotheses: the <real line> is <closed> in itself but not <compact>; in a <compact space>, every <closed subset> is <compact>.
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