A subset E of $R^n$ is totally bounded if and only if E is bounded

I am studying Compactness in metric space with Gamelin and Greene's Introduction to Topology and am confused about lemma 5.4 in the book.

A metric space $X$ is totally bounded if for each $e > 0$, there exists a finite number of open balls of radius $e$ that covers $X$.

Lemma 5.4

A subset $E$ of $\Bbb R^n$ is totally bounded if and only if $E$ is bounded.

To prove the reverse direction, $E$ is contained in a cube of the form $T = [-b,b] \times\dots \times [-b,b]$ for some large $b > 0$. Then, since any subspace of a totally bounded metric space is totally bounded, it suffices to show that T is totally bounded. For this, let $e > 0$. Then the balls $B(e j;e)$ cover $T$, where $j = (j_1,...,j_n)$ ranges over all integral lattice points of $R^n$ which satisfy $e|j_i| \le 2b$, where $i$ ranges from $1$ to $n$. Since there are only finitely many such lattice points, $T$ is totally bounded.

I don't understand the bold part of the proof. I guess they created an arbitrary e-ball by putting the center of the ball as the scalar multiplication of j by e. But then I don't see how the restiction of lattice points which satisfy, $e|j_i| \le 2b$, make makes the balls cover $T$.

Can anyone clear me up? Thanks.