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

Proof
We use the following lemma.

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

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

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

For $m \in \Z_{>0}$, we apply the above lemma to $\sequence {x^{\paren {m - 1} }_n}_{n \mathop \in \N}$.

Then:
 * $\exists c_m \in A$

and:
 * $\sequence {x^{\paren {m - 1} }_n}_{n \mathop \in \N} \supseteq \exists \sequence {x^{\paren m}_n}_{n \mathop \in \N}$ a subsequence such that:
 * $\forall n \in \N: \map d {x^{\paren m}_n, c_m} < \dfrac 1 m$

Note that the above procedure to define the infinitely nested subsequences is justified by the axiom of countable choice.

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

Finally, we shall show $\sequence {y_n}_{n \mathop \in \N}$ is convergent.

As $\struct {A, d}$ is complete, it suffices to show it is a Cauchy sequence.

Let $\epsilon \in \R_{>0}$ be arbitrary.

Let $N \in \N$ so large that:
 * $\dfrac 2 N < \epsilon$

Then, by :
 * $\forall m, n \ge N : \map d {y_m, y_n} \le \map d {y_m, c_N} + \map d {y_n, c_N} < \dfrac 1 N + \dfrac 1 N < \epsilon$