 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 $U_{y_1},\dots ,U_{y_r}$ 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 $K$ is closed. The empty compact set is closed as well.