 Solution

Let $K$ be a compact set of a Hausdorff space $X$. For any $x\notin K$ and each $y\in K$, the Hausdorff property gives disjoint open sets $U_y\ni y$ and $V_y\ni x$. The sets $U_y$ cover $K$; compactness supplies a finite subcover $U_{y_1},\dots ,U_{y_r}$. The open set $V={\bigcap }_i{V}_{y_i}$ contains $x$ and misses every $U_{y_i}$, hence misses $K$. Thus $X\setminus K$ is open, and every compact subset of a Hausdorff space is closed. The empty-set case is immediate.

Since $\mathbb{R}$ with its usual topology is Hausdorff, its compact sets are closed. The open cover $\{(-n,n):n\ge 1\}$ of $\mathbb{R}$ has a finite subcover of any compact set $K$. Taking the largest index in that subcover gives $K\subset (-N,N)$, so $K$ is a bounded set. This proves the required direction of the Heine-Borel theorem without assuming its converse.