= Solution
Let $K$ be a compact subset of a <Hausdorff space>, and take $x\notin K$. For each $y\in K$, the definition of a <Hausdorff space> supplies disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$. Compactness gives finitely many $U_{y_1},\ldots,U_{y_r}$ covering $K$. Their corresponding intersection $V=\bigcap_{j=1}^r V_{y_j}$ is open, contains $x$, and misses $K$. Therefore every point of the complement has an open neighbourhood contained in the complement, so \b[$K$ is closed]. The empty compact set is closed as well.
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