Sequentially Compact Metric Space is Compact/Proof 2

Proof
Let $M = \left({A, d}\right)$ be a sequentially compact metric space.

Let $\mathcal U$ be any open cover of $M$.

By Lebesgue's Number Lemma, there exists a Lebesgue number for $\mathcal U$.

By Sequentially Compact Metric Space is Totally Bounded, there exists $\left\{{x_1, x_2, \ldots, x_n}\right\}$ which is a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number.

Let $B_\epsilon \left({x_i}\right)$ be the open $\epsilon$-ball of $x_i$.

By definition of Lebesgue number, $B_\epsilon \left({x_i}\right)$ is contained in some $U_i \in \mathcal U$.

Since:
 * $\displaystyle M \subseteq \bigcup_{i \mathop = 1}^n B_\epsilon \left({x_i}\right) \subseteq \bigcup_{i \mathop = 1}^n U_i$

we have a finite subcover $\left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal{U}$ for $M$.

Hence the result.