Complete and Totally Bounded Metric Space is Sequentially Compact/Proof 2

Proof
The results:
 * Compact Space is Countably Compact
 * Countably Compact Metric Space is Sequentially Compact

show that it suffices to prove that $M$ is compact.

Using a Proof by Contradiction, assume that $M$ is not compact.

Let $\mathcal C$ be an open cover for $A$ such that $\mathcal C$ does not have a finite subcover for $A$.

By the definition of a totally bounded metric space, we can use the axiom of countable choice to obtain a sequence $\left\langle{F_n}\right\rangle_{n \in \N}$ such that:
 * For all $n \in \N$, $F_n$ is a finite $2^{-n}$-net for $M$.

For the sake of notational brevity, let $B'_n \left({a}\right)$ denote the open $2^{-n}$-ball of $a$ in $M$.

For all $n \in \N$, define:
 * $G_n = \left\{{y}\right. \in F_n: B'_n \left({y}\right)$ is not covered by any finite subset of $\left.{\mathcal C}\right\}$

Since $F_n$ is finite by definition, it follows by the definition of a net that $G_n$ is non-empty.

For all $y \in G_n$, define:
 * $T_n \left({y}\right) = \left\{{z}\right. \in G_{n+1}: B'_n \left({y}\right) \cap B'_{n+1} \left({z}\right)$ is not covered by any finite subset of $\left.{\mathcal C}\right\}$

By the definition of a net, it follows from the distributivity of intersection over union that:
 * $\displaystyle B'_n \left({y}\right) = \bigcup_{z \mathop \in F_{n+1}} \left({B'_n \left({y}\right) \cap B'_{n+1} \left({z}\right)}\right)$

Hence, by the definition of $G_n$, it follows that $T_n \left({y}\right)$ is non-empty.

Since the countable union of countable sets is countable, it follows that the disjoint union $\displaystyle \bigsqcup_{n \mathop \in \N} G_n$ is countable.

Using the axiom of countable choice, we can obtain a sequence $\left\langle{\phi_n: G_n \to G_{n+1}}\right\rangle_{n \in \N}$ of mappings such that:
 * $\forall n \in \N: \forall y \in G_n: \phi_n \left({y}\right) \in T_n \left({y}\right)$

Let $x_0 \in G_0$.

For all $n \in \N$, define:
 * $x_{n+1} = \left({\phi_n \circ \cdots \circ \phi_1 \circ \phi_0}\right) \left({x_0}\right)$

where $\circ$ denotes composition of mappings.

For all $n \in \N$, define:
 * $A_n = B'_n \left({x_n}\right) \cap B'_{n+1} \left({x_{n+1}}\right)$

Note that $A_n$ is non-empty; otherwise, by Union of Empty Set, $\varnothing$ would be a cover for $A_n$.

Let $y \in A_n$. Then:
 * $\displaystyle d \left({x_n, x_{n+1}}\right) \le d \left({x_n, y}\right) + d \left({x_{n+1}, y}\right) < \frac 1 {2^n} + \frac 1 {2^{n+1}} = \frac 3 {2^{n+1}}$

Let $i < j$, where $i, j \in \N$.

Then, by Sum of Infinite Geometric Progression:
 * $\displaystyle d \left({x_i, x_j}\right) \le \sum_{k \mathop = i}^{j-1} d \left({x_k, x_{k+1}}\right) < \sum_{k \mathop = i}^{\infty} \frac 3 {2^{k+1}} = \frac 3 {2^i}$

Hence, by Power of Number less than One, the sequence $\left\langle{x_k}\right\rangle$ is Cauchy.

By the assumption that $M$ is complete, the sequence $\left\langle{x_k}\right\rangle$ converges to some limit $x \in A$.

Choose $U \in \mathcal C$ such that $x \in U$ (which can be done because $\mathcal C$ covers $A$).

By the definition of an open set, we can choose a strictly positive real number $\epsilon$ such that $B_{\epsilon} \left({x}\right) \subseteq U$.

By Power of Number less than One, we can choose a natural number $n$ such that:
 * $\displaystyle \frac 1 {2^n} < \frac \epsilon 2$

By the definition of a limit, we can choose a natural number $m > n$ such that:
 * $\displaystyle d \left({x_m, x}\right) < \frac 1 {2^n}$

For all $y \in A_m$, we have:
 * $\displaystyle d \left({y, x}\right) \le d \left({y, x_m}\right) + d \left({x_m, x}\right) < \frac 1 {2^m} + \frac 1 {2^n} < \epsilon$

That is, $A_m \subseteq B_{\epsilon} \left({x}\right) \subseteq U$.

Since $\subseteq$ is a transitive relation, it follows that $A_m \subseteq U$.

That is, $A_m$ is covered by the singleton $\left\{{U}\right\} \subseteq \mathcal C$.

But this contradicts the definitions of $T_m \left({x_m}\right)$ and $\phi_m$.