Complete and Totally Bounded Metric Space is Sequentially Compact/Lemma

Lemma
Let $\struct { A, d}$ be a totally bounded metric space.

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

Then:
 * $\forall r\in\R _{>0}\;\exists c\in A: \map d {x_n, c} < r$ for infinitely many $n$.

Proof
Since $\struct {A, d}$ is totally bounded, there are $c_0,\ldots, c_k\in A$ such that:
 * $\forall x\in A: \inf _{i \in \closedint 0 k}\map d {x, c_i} < r$

In particular:
 * $\N = \bigcup _{i \in \closedint 0 k} \set {n\in\N : \map d {x_n, c_i} < r}$

As $\size \N = \infty$, there exists an $i \in \closedint 0 k$ such that:
 * $\size {\set {n\in\N : \map d {x_n, c_i} < r}} = \infty$