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
 * $$\inf_{0\leq i\leq n}d(x_i,x)\leq 1$$

for all $$x\in S$$. Let us set $$a := x_0$$,
 * $$ D := \max_{0\leq i\leq n} d(x_0,x_i)$$

and $$K := D + 1$$. 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.