Totally Bounded Metric Space is Second-Countable/Proof 1

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

Then $M$ is second-countable.

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

Consider the following set of open balls:
 * $\mathcal C := \left\{{N_{1/k} \left({x}\right) : k \in \N^*}\right\}$

As $M$ is totally bounded, each one forms an $\epsilon$-net where $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$

Hence, $\mathcal C$ is a countable basis for $X$.

That is, $M$ is second-countable.