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
For all $n \in \N$, let:
 * $\mathcal C_n = \left\{ {B_{1/n}\left({x}\right) : x \in X} \right\}$

where $B_{\epsilon}\left({x}\right)$ denotes the open $\epsilon$-ball of $x$ in $X$.

By the definition of total boundedness, for all $n \in \N$, there exists a finite subcover $\mathcal F_n$ of $\mathcal C_n$ for $X$.

Define:
 * $\displaystyle \mathcal B = \bigcup_{n \mathop \in \N} \mathcal F_n$

Clearly, $\mathcal B$ is countable.

It suffices to show that $\mathcal B$ is a basis for the topology induced by the metric $d$.

Let $U$ be any open subset of $X$.

For all $x \in U$, there exists an $n \in \N$ such that $N_{2/n}\left({x}\right) \subseteq U$.

Because $\mathcal F_n$ covers $X$, there exists a $B \in \mathcal F_n$ such that $x \in B$.

Let $c$ be the center of the open $\epsilon$-ball $B$. Recall that $B$ has radius $\dfrac 1 n$ by the definition of $\mathcal F_n$.

For any $y \in B$:
 * $\displaystyle d \left({x, y}\right) \le d \left({x, c}\right) + d \left({y, c}\right) \le \frac 2 n$

Therefore, $y \in N_{2/n} \left({x}\right)$.

Hence $B \subseteq N_{2/n} \left({x}\right) \subseteq U$.

We have just shown that for all $x \in U$, there exists an $n \in \N$ and a $B \in \mathcal F_n$ such that $x \in B \subseteq U$.

Hence:
 * $\displaystyle U \subseteq \bigcup_{n \mathop \in \N} \bigcup \left\{ {B \in \mathcal F_n : B \subseteq U} \right\}$

It is also true that:
 * $\displaystyle U \supseteq \bigcup_{n \mathop \in \N} \bigcup \left\{ {B \in \mathcal F_n : B \subseteq U} \right\}$

It follows that:
 * $\displaystyle U = \bigcup_{n \mathop \in \N} \bigcup \left\{ {B \in \mathcal F_n : B \subseteq U} \right\}$

This is a union of members of $\mathcal B$.

Hence $\mathcal B$ is a basis for the topology induced by the metric $d$.

Therefore, $X$ is second-countable.