Complete and Totally Bounded Metric Space is Sequentially Compact/Proof 4

Proof
We use the following lemma, that depends on the axiom of countable choice.

Lemma
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $A$.

In the following, we will construct recursively nested subsequences:
 * $\sequence {x^{\paren 0}_n}_{n \mathop \in \N}\supseteq \sequence {x^{\paren 1}_n}_{n \mathop \in \N}\supseteq \sequence {x ^{\paren 2}_n}_{n \mathop \in \N} \supseteq\cdots$

such that:
 * $\forall N\in\Z_{>0} : \forall k,l \in \N: \map d {x^{\paren N}_k, x^{\paren N}_l} \le \dfrac 1 N$

First, let:
 * $\sequence {x^{\paren 0}_n}_{n \mathop \in \N} = \sequence {x_n}_{n \mathop \in \N}$

Then, for each $N \in \Z_{>0}$, applying $f$ from lemma, let:
 * $\sequence {x^{\paren N}_n}_{n \mathop \in \N} := \map f {N, \sequence {x^{\paren {N - 1} }_n}_{n \mathop \in \N}}$

Now, let:
 * $y_n := x^{\paren n}_n$

for all $n \in \N$.

By construction:
 * $\sequence {y_n}_{n \mathop \in \N}$ is a subsequence of $\sequence {x_n}_{n \mathop \in \N}$

and:
 * $\forall N \in \Z_{>0}: \sequence {y_n}_{n \mathop \ge N}$ is a subsequence of $\sequence {x^{\paren N}_n}_{n \mathop \in \N}$

From the latter follows:
 * $\forall N\in\Z_{>0} : \forall m, n \ge N : \map d {y_m, y_n} \le \dfrac 1 N $

that implies that $\sequence {y_n}_{n \mathop \ge N}$ is a Cauchy sequence.

As $\struct {A, d}$ is complete, $\sequence {y_n}_{n \mathop \in \N}$ is convergent.