Compact sets are closed and bounded

Closed. If $x_n\in K$ converges to $x$, sequential compactness gives a convergent subsequence in $K$, so $x\in K$; hence $K$ is closed.

Bounded. If $K$ were unbounded, there would exist $(x_n)$ in $K$ with $d(x_0,x_n)\to+\infty$; such a sequence has no convergent subsequence, contradiction.

Conclusion. In metric spaces, compact $\Rightarrow$ closed and bounded.

Note. The converse holds in $\mathbb{R}^n$ (Heine–Borel) but not in general metric spaces.