Totally Bounded Metric Space is Separable

Theorem
A totally bounded metric space is separable.

Proof
Let $M = \left({X, d}\right)$ be a totally bounded metric space.

By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\left\langle{F_n}\right\rangle_{n \ge 1}$ such that:
 * For all natural numbers $n \ge 1$, $F_n$ is a finite $\left({1/n}\right)$-net for $M$.

Let $\displaystyle S = \bigcup_{n \mathop \ge 1} F_n$.

Since the countable union of countable sets is countable, it follows that $S$ is countable.

It suffices to prove that $S$ is everywhere dense in $M$.

Let $S^-$ denote the closure of $S$.

Let $x \in X$.

Let $U \subseteq X$ be open in $M$ such that $x \in U$.

By definition, there exists a strictly positive real number $\epsilon$ such that $B_{\epsilon} \left({x}\right) \subseteq U$; i.e., the open $\epsilon$-ball of $x$ in $M$ is contained in $U$.

By the Archimedean principle, there exists a natural number $\displaystyle n > \frac 1 \epsilon$.

That is, $\displaystyle \frac 1 n < \epsilon$, and so $B_{1/n} \left({x}\right) \subseteq B_{\epsilon} \left({x}\right)$.

Since $\subseteq$ is a transitive relation, we have $B_{1/n} \left({x}\right) \subseteq U$.

By the definition of a net, there exists a $y \in F_n$ such that $x \in B_{1/n} \left({y}\right)$.

By axiom $\left({M3}\right)$ for a metric, it follows from the definition of an open ball that $y \in B_{1/n} \left({x}\right)$.

Since $y \in S \cap U$, it follows that $x$ is an adherent point of $S$.

By Equivalence of Definitions of Adherent Point, we have $x \in S^-$.

That is, $X \subseteq S^-$, and so $S$ is everywhere dense in $M$.