Sequentially Compact Metric Space is Totally Bounded/Proof 1

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

Let $M$ be sequentially compact.

Then $M$ is totally bounded.

Proof
Let $\epsilon \in \R_{>0}$.

Proof by contradiction.

By induction, $\forall n \in \N_{>0}: \mathcal S_n = \left\{{S \subseteq A: \left\vert{S}\right\vert = n, \, \forall x, y \in S: x \ne y \implies d \left({x, y}\right) \ge \epsilon}\right\}$ is non-empty.

ACC gives sequence $\left\langle{F_n}\right\rangle_{n \ge 1}$ such that $F_n \in \mathcal S_n$ for all natural numbers $n \ge 1$.

Let $\displaystyle \phi: \bigcup_{n \mathop \ge 1} F_n \to \N$ be an injection (possible by Countable Union of Countable Sets is Countable).

Let $x_1 \in F_1$.

Assume $x_1, x_2, \ldots, x_n$ are already defined.

Let $x_{n+1} \in F_{n+1}$ be the unique element of $A$ such that:
 * $\displaystyle \forall k \in \N: 1 \le k \le n \implies d \left({x_{n+1}, x_k}\right) \ge \frac \epsilon 2$ $(*)$
 * $\phi \left({x_{n+1}}\right)$ is smallest possible

Uniqueness follows from injectivity of $\phi$.

Proof of existence of $y \in F_{n+1}$ satisfying $(*)$ (Proof by contradiction):

Suppose $\displaystyle \forall y \in F_{n+1}: \exists k \in \N: 1 \le k \le n: d \left({y, x_k}\right) < \frac \epsilon 2$.

By Finite Cartesian Product of Non-Empty Sets is Non-Empty, let $f: F_{n+1} \to \left\{{x_0, x_1, \ldots, x_n}\right\}$ such that $\displaystyle d \left({y, f \left({y}\right)}\right) < \frac \epsilon 2$.

By the Pigeonhole Principle, $\exists y, z \in F_{n+1}: y \ne z: f \left({y}\right) = f \left({z}\right)$.

Then $d \left({y, z}\right) \le d \left({y, f \left({y}\right)}\right) + d \left({z, f \left({z}\right)}\right) < \epsilon$, contradicting definition of $F_{n+1}$.

Then $\left\langle{x_n}\right\rangle$ satisfies:
 * $\displaystyle \forall m, n \in \N: m \ne n \implies d \left({x_m, x_n}\right) \ge \frac \epsilon 2$

and therefore has no Cauchy subsequence (Contradiction).