Totally Bounded Metric Space is Second-Countable/Proof 2

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

Then $M$ is second-countable.

Proof
Using the axiom of countable choice, we can apply the definition of a totally bounded metric space to conclude that there exists 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 $M$.

Define:
 * $\displaystyle \mathcal B = \bigcup_{n \mathop \ge 1} \left\{{B_{1/n} \left({x}\right): x \in F_n}\right\}$

Sicne the countable union of countable sets is countable, it follows that $\mathcal B$ is countable.

Let $\tau$ be the topology on $X$ induced by the metric $d$.

It suffices to show that $\mathcal B$ is an analytic basis for $\tau$.

Since an open ball is open, we have that $\mathcal B \subseteq \tau$.

We use Equivalent Definitions of Analytic Basis.

Let $y \in U \in \tau$.

By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $B_{\epsilon} \left({y}\right) \subseteq U$.

By the Archimedean principle, there exists a natural number $\displaystyle n > \frac \epsilon 2$.

That is, $\displaystyle \frac 2 n < \epsilon$, and so $B_{2/n} \left({y}\right) \subseteq B_{\epsilon} \left({y}\right)$.

Since $\subseteq$ is a transitive relation, we have $B_{2/n} \left({y}\right) \subseteq U$.

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

For all $z \in B_{1/n} \left({x}\right)$, we have:

That is, $B_{1/n} \left({x}\right) \subseteq B_{2/n} \left({y}\right)$.

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

Hence the result.