Totally Bounded Metric Space is Separable

Theorem
A totally bounded metric space is separable and second-countable.

Proof
Consider a totally bounded metric space $(X,\vartheta)$.

Let us first show that $(X,\vartheta)$ is separable, that is, that we can find a countable dense set in $X$.

As we can cover $X$ by a finite number of balls of any given size, let us define a set of points as follows:
 * $X$ can be covered by a certain finite number of balls of radius $1$; call their centers $x_{1,1},x_{1,2},\ldots,x_{1,n_1}$, for some natural number $n_1$.
 * $X$ can be covered by a certain finite number of balls of radius $1/2$; call their centers $x_{2,1},x_{2,2},\ldots,x_{2,n_2}$, for some natural number $n_2$.
 * In general, for an integer $k \geq 1$, $X$ can be covered by a certain finite number of balls of radius $1/k$; call their centers $x_{k,1},x_{k,2},\ldots,x_{k,n_k}$, for some natural number $n_k$.

Consider the set of all of these centers:
 * $S := \{x_{k,n} \mid k,n \in \N, \, 1 \leq k \leq n_k \}$.

This set fulfills the requirements: it is countable, as it is a countable union of finite sets. It also dense in $X$, as we will prove now:

Take any point $x \in X$, and any $\epsilon > 0$, and let us prove that $B(x,\epsilon) \cap S \neq \emptyset$.

Take some natural $k$ such that $1/k < \epsilon$.

As the balls of centers $x_{k,1}, x_{k,2},\ldots,x_{k,n_k}$ and radius $1/k$ cover $X$, there exists a ball, say $B(x_{k,n}, 1/k)$, which contains $x$, the center of our initial ball.

But then:
 * $d(x_{k,n}, x) < 1/k < \epsilon,$

which implies that $x_{k,n} \in B(x,\epsilon)$.

The point $x_{k,n}$ is also in $S$, so we have proved that $S$ is dense in $X$.

Then we have that a Totally Bounded Metric Space is Second-Countable.