Totally Bounded Metric Space is Second-Countable/Proof 1

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

Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$

As $M$ is totally bounded, for each $\epsilon$ there exists a finite $\epsilon$-net $\mathcal C$ for $M$.

From Net forms Basis for Metric Space, $\mathcal C$ is a countable basis for $M$.

That is, $M$ is second-countable.