If every closed ball in a metric space $X$ is compact, show that $X$ is separable.

Question

The problem says:

If every closed ball in a metric space $X$ is compact, show that $X$ is separable.

I'm trying to use an equivalence in metric spaces that tells us: let X be the matric space, the following are equivalent

X is 2nd countable

X is Lindeloff

X is separable

I also thought about taking balls of a "big" radius and that they are disjoint, but I do not see how to make the set of balls is countable.

Answer 1

$X = \cup_{n \in \mathbb{N}} D(p, n)$, for any $p \in X$ (where $D(x,r)$ denotes the closed ball around $p$ of radius $r$).

So $X$ is $\sigma$-compact hence Lindelöf hence separable.

Answer 2

For a different approach, choose $x\in X$ so that $X = \cup_{n \in \mathbb{N}} \overline B_x(n).$ Now consider, for fixed $n\in \mathbb N,$ the fact that the compact ball $\overline B_x(n)$ is totally bounded. This means that, for each $m\in \mathbb N$, there is a $\textit{finite}$ set of points $x_{m,n}\in \overline B_x(n)$ such that if $y\in \overline B_x(n)$ then there is an $m\in \mathbb N$ and $x_{m,n}$ such that $d(y,x_{m,n})<1/m,$ which implies that $\left \{ x_{m,n} \right \}_{m\in \mathbb N}$ is a countable dense subset of $\overline B_x(n)$ and therefore that $\left \{ x_{m,n} \right \}_{m,n\in \mathbb N}$ is a countable dense subset of $X$. The result follows by the equivalence separable $\Leftrightarrow $ second countable.