An elementary way to show any bounded subset of $\Bbb{R}^k$ is totally bounded I'm trying to show that any subset bounded of $\Bbb{R}^k$ is totally bounded.
Here is what I did: 
(1)A subset of a totally bounded Set is bounded:
Proof: Let $X$ be a totally bounded subset and $Y\subset X$ then there exists an $\epsilon /2$-net $\{x_1,x_2,..,x_n\}$ and $X\subset \displaystyle\bigcup_{i=1}^n B(x_i,\epsilon/2)$ . Let $\{x_1,x_2,..,x_m\}$ be the points whose balls contain $Y$ $(m\le n)$. Now $\forall i \in \{1,..,m\} \exists q_i \in A \cap B(x_i,\epsilon/2) $ and $B(x_i,\epsilon/2)\subset B(q_i,\epsilon)$. We have for every $x\in B(x_i,\epsilon/2)$ $$ d(x,q_i)\le d(x,x_i)+d(x_i,q_i) < \frac \epsilon 2 + \frac \epsilon 2 =\epsilon$$
Hence $Y \subset \displaystyle\bigcup_{i=1}^m B(x_i,\epsilon/2) \subset \bigcup_{i=1}^m B(q_i,\epsilon)$ and $q_i \in Y$ for all $i$ hence $Y$ is totally bounded
Back to the problem:
Let $A \subset \Bbb{R}^k$ be a bounded set  then $A \subset B(0,R)$ for some $R$ then $A \subset [-R,R] \times [-R,R] \times ...\times [-R,R] $ then $A$ is a subset of a compact set by the Heine-Borel Theorem which is also a totally bounded set, hence by (1) $A$ is totally bounded.
I'm trying to do this problem by only using (1) without invoking the Heine-Borel theorem, can anyone tell me how that can be done? (Is the proof of (1) right for that matter?)
 A: If $B$ is bounded in $\mathbb R^n$ it is contained in some cube $[-R,R]^n$. Then for any $\epsilon>0$ just consider all balls of radius $\epsilon$ centered in the set
$$
  (\frac \epsilon 2\mathbb Z)^n \cap [-R,R]^n
$$
these balls are a finite number (less then $(4R/\epsilon+1)^n$) and cover the whole cube $[-R,R]^n(edit this)$.
A: I'll be systematic here, I think it can help.
D Let $S$ be any subset of $\Bbb R^n$. Given $\epsilon >0$, we say that $N$ is an $\epsilon$-net for $S$ if the set of open balls
$$B_\epsilon(N)=\{B(x,\epsilon):x\in N\}$$
covers $S$. That is, the set of open balls of radius $\epsilon$ centered at the points of $N$ cover $S$. 
D We say a subset $S$ of $\Bbb R^n$ is precompact or totally bounded if for every $\epsilon >0$; there exists a finite $\epsilon$-net for $S$.
T Let $S$ be bounded in $\Bbb R^n$. Then $S$ is precompact.
P Boundedness implies $S$ is contained in some closed ball $B$. But each of these balls contain but a finite number of elements of the form 
$${\bf x}_{\ell,{\bf k}}= \left(\frac{k_1}{2^\ell},\dots,\frac{k_n}{2^\ell}\right)$$ for $k_i,\ell \in \Bbb Z\;\;;\ell \geq 0$, a fixed number, while the $k_i$ varies independently through the integers. But then, given $\epsilon >0$, we can take $\ell $ sufficiently large so that $\frac 1 {2^\ell}<\epsilon$, and the set of such points ${\bf x}_{\ell,k}$ contained in $B$ will be a finite $\epsilon$-net for $S$.
NOTE Observe the proof simply relies in producing what we usually think a net is: we show the intersection of our ball with the grid of "mesh" $1/2^\ell$ is finite, and then show that this intersection is an $\epsilon$-net (since we make $2^{-\ell}$ small) of the underlying set $S$ inside $B$. Note that we use the $k$ in the denominator the eventually "get out" of the open ball (since it is bounded, some natural will make $k/2^\ell$ "leave" the ball, no matter how small $2^{-\ell}$ is.)
