a compact set in a Hausdorff space is closed

Theorem. A compact set in a Hausdorff space is closed.

Proof.Let $A$ be a compact set in a Hausdorff space $X$.The case when $A$ is empty is trivial, so let usassume that $A$ is non-empty.Using this theorem (http://planetmath.org/APointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods),it follows that each point$y$ in $A^{\mathrm{\complement }}$ has a neighborhood $U_y$, whichis disjoint to $A$. (Here, we denote the complement of $A$by $A^{\mathrm{\complement }}$.)We can therefore write

$A^{\mathrm{\complement }}$$=$${\displaystyle \bigcup _{y\in A^{\mathrm{\complement }}}}{U}_y.$

Since an arbitrary union of open sets is open, it follows that $A$ isclosed. $\mathrm{\square }$

Note. 
The above theorem can, for instance, be found in [1] (page 141),or [2] (Section 2.1, Theorem 2).