Totally Bounded Metric Space is Bounded

Lemma
A totally bounded metric space $(S,d)$ is bounded.

Proof
Suppose that $(S,d)$ is totally bounded.

Then there exist $n \in \N$ and points $x_0, \dots, x_n\in S$ such that:
 * $\displaystyle \inf_{0\leq i\leq n}d(x_i,x)\leq 1$

for all $x\in S$.

Let us set:
 * $a := x_0$
 * $\displaystyle D := \max_{0 \leq i \leq n} d(x_0,x_i)$
 * $K := D + 1$.

Now let $x \in S$ be arbitrary.

Then by assumption there exists $i$ such that $d(x_i,x)\leq 1$.

Hence:
 * $ d(a,x) \leq d(a,x_i) + d(x_i,x) \leq 1 + D = K$

So $(S,d)$ is bounded, as claimed.