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 us assume that $A$ is non-empty. Using this theorem, it follows that each point $y$ in $A^{\comp}$ has a neighborhood $U_y$ , which is disjoint to $A$ . (Here, we denote the complement of $A$ by $A^{\comp}$ .) We can therefore write \begin{eqnarray*} A^{\comp} &=& \bigcup_{y\in A^{\comp}} U_y. \end{eqnarray*}Since an arbitrary union of open sets is open, it follows that $A$ is closed.

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

Bibliography

1

J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.

2

I.M. Singer, J.A.Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.