Sequentially Compact Metric Space is Compact/Proof 2

Theorem
A sequentially compact metric space is compact.

Proof
Let $M = \left({X, d}\right)$ 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 has Finite Net, 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.

Then $N_{\epsilon} \left({x_i}\right)$ is contained in some $U_i \in \mathcal{U}$ by definition of Lebesgue number.

Since $\displaystyle M \subseteq \bigcup_{i=1}^n N_{\epsilon} \left({x_i}\right) \subseteq \bigcup_{i=1}^n U_i$, we have a finite subcover $\left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal{U}$ for $M$.

Hence the result.