The converse of the Heine–Borel property I have been reading on the Heine–Borel theorem and Heine–Borel property and their relation to topological vector spaces.
The Heine–Borel theorem states each subset of Euclidean space $\mathbb{R}^n$, is closed and bounded if and only if it is compact.
A topological vector space is said to have the Heine–Borel property if each closed and bounded set in is compact.
From this I understand that not every TVS has the Heine–Borel property. However, what about the converse? i.e, Is each compact subset of TVS, closed and bounded?

In a topological vector space, one point sets (which are clearly compact) are closed if and only if the space is Hausdorff (see, for example, Rudin's book on functional analysis). In a Hausdorff space, every compact set is closed.
If $C$ is compact and $V$ an open neighborhood of $0$, the family of sets of the form $nV$ is an open cover of $C$. If a finite subcover exists, there must exist a cover given by a single such $n V$, so every compact set in a topological vector space is bounded.
