Let $E,F$ be normed spaces, $S,T \in L(E,F)$ such that T is compact and S is not. Let $M \subset E$ be a bounded set. By definition $\overline{T(M)} $ is compact. Now is it true that the set $T(M)$ is totally bounded while $S(M)$ is not? Since the following holds:
$A$ is compact $\iff$ $A$ is totally bounded and closed .
Since we take the closure of $T(M)$ it must be true that $T(M)$ is totally bounded. So it must be possible to find a bounded linear map that takes a bounded set to a set that is not totally bounded. How is that possible?
The existence of a bounded set in a normed vector space which is not totally bounded is sufficient, because then we can consider $F=E$ and $S$ the identity map on $E$.
So consider $E=\ell^2$, and let $B$ be the closed unit ball of $\ell^2$, i.e., $B=\{x\in\ell^2:\|x\|\leq1\}$. Then $B$ is clearly bounded, but it is not totally bounded. For it is closed, and if it were totally bounded, it would then be compact, which is not true.