The first simplification is to consider $r=1$. The second simplification is to consider $n=2$, and the third, $x=0$. So, what's the box of side-length $2$ containing the unit ball, i.e. unit disk, in $\Bbb{R}^2$? For this one, you can just draw a picture.
Increasing the dimension is probably the apparently trickiest bit. But the reason you could put the unit disk $D^2$ into the square (spoilers!) $[-1,1]\times[-1,1]$ is that $\inf\{x:(x,y)\in D^2\}=\inf\{x:(x,y)\in D^2\}-1$, and similarly for the suprema in each coordinate. In other words, for every $(x,y)\in D^2, -1\leq x\leq 1$ and $-1\leq y \leq 1$. But that's saying exactly that $D^2\subseteq [-1,1]\times [-1,1]$! In almost exactly the same way you can get $B_1(0)\subseteq [-1,1]^n$ for every $n$. Now can you generalize to all $r$ and $x$?
Once you've finished this, there's nothing to your second question: a bounded set is by definition one that lies in a ball, and you'll have just shown that every ball lies in a box, even with a very explicit bound on the size of the box.