= Solution
Let $K\subseteq X$ be <compact>, with $X$ a <Hausdorff space>. The <empty set> is <closed>, so suppose $K\neq\varnothing$. Fix $x\in X\setminus K$. For each $y\in K$, the <Hausdorff> property supplies disjoint <open> <neighbourhoods> $U_y$ of $y$ and $V_y$ of $x$.
The $U_y$ form an <open cover> of $K$. Choose a <finite subcover> $U_{y_1},\ldots,U_{y_m}$ and put $V=\bigcap_{r=1}^m V_{y_r}$. This is an <open> <neighbourhood> of $x$. If a point of $V$ belonged to $K$, it would belong to some $U_{y_r}$ as well as $V_{y_r}$, a contradiction. Hence $V\subseteq X\setminus K$. Every point of the complement has such an <open> <neighbourhood>, so $X\setminus K$ is <open> and $K$ is <closed>.
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