An elementary way to show any bounded subset of $\Bbb{R}^k$ is totally bounded

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 $[-R,R]^n(edit this)$.

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.)