Euclidean bounded metric space implies total boundedness How to prove that an Euclidean bounded metric space is also totally bounded? 
 A: Let $M$ be a bounded subspace of $\Bbb R^n$. For $r>0$ let $K_r=\{x\in\Bbb R^n:\|x\|\le r\}$. Since $M$ is bounded, there is an $r>0$ such that $M\subseteq K_r$. Fix $\epsilon>0$. $K_r$ is totally bounded, so there is a finite $F\subseteq K_r$ such that $$M\subseteq K_r\subseteq\bigcup_{x\in F}B\left(x,\frac{\epsilon}2\right)\;.$$
Let $F_0=\left\{x\in F:B\left(x,\frac{\epsilon}2\right)\cap M\ne\varnothing\right\}$, for each $x\in F_0$ pick a point $\hat x\in B\left(x,\frac{\epsilon}2\right)\cap M$, and let $F_1=\left\{\hat x:x\in F_0\right\}$. Note that if $y\in B\left(x,\frac{\epsilon}2\right)$, then $$\|\hat x-y\|\le\|\hat x-x\|+\|x-y\|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon\;,$$ so $B\left(x,\frac{\epsilon}2\right)\subseteq B\left(\hat x,\epsilon\right)$, and $F_1$ is an $\epsilon$-net for $M$.
A: A simple proof that relies on a few well-know results about $\mathbb{R}^n$: 
Firstly, $(X,d)$ is totally bounded if and only if every sequence has a Cauchy sub-sequence. So if $A \subseteq \mathbb{R}^n$ is bounded, then every sequence $(x_n)$ is bounded, and therefore has a convergent subsequence - which is Cauchy. 
