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 $c_0, \dots, c_k \in A$ such that:
 * $\ds \forall x \in A: \inf_{i \mathop \in \closedint 0 k} \map d {x, c_i} < r$

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

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