Are boundedly compact metric space always equal to a countable union of their compact subsets

Question

We say a metric space is boundeldy compact if every closed and bounded subset of it is compact.

Question: Is the following true? "Any boundedly compact metric space is a countable union of its compact subsets."

Answer 1

Let $X$ be boundedly compact and take $x \in X$.

Since the closed balls $\overline{B}(x; 2^n)$, for $n \in \mathbb{Z}$, are closed and bounded, they're compact. But, $X = \cup_{n \in \mathbb{Z}} \overline{B}(x; 2^n)$. Therefore the claim follows.