Theorem 1   Every totally bounded subset of a metric space is bounded.

Proof. Let $K$ be a totally bounded subset of a metric space. Suppose $x,y \in K$ . We will show that there is $M>0$ such that $d(x,y)<M$ . From the definition of totally bounded, we can find an $\varepsilon>0$ and a finite subset $\{x_1,x_2...,x_n\}$ of $K$ such that $K\subseteq \bigcup_{k=1}^n B(x_k,\varepsilon )$ , so $x\in B(x_i,\varepsilon )$ ,$y\in B(x_l,\varepsilon)$ , $i,l \in \{1,2,...n\}$ . So we have that \begin{eqnarray*} d(x,y) & \leq & d(x,x_i)+d(x_i,x_l)+d(x_l,y)\\ & < & \varepsilon +\max_{1\leq{s,t}\leq n}d(x_s,x_t)+\varepsilon = M \end{eqnarray*}