Let $X,Y$ be normed vector spaces and $A:X\to Y$ be linear and such that $AB^X_1(0)$ is relatively compact (where $B^X_1(0)$ denotes the open unit ball in $X$).
Why can we conclude that $A$ is bounded?
Since $AB^X_1(0)$ is relatively compact, we know that $\overline{AB^X_1(0)}$ is totally bounded and hence bounded. So, there is a $c\ge0$ with $$\sup_{x\in B^X_1(0)}\left\|Ax\right\|\le\sup_{y\in\overline{AB^X_1(0)}}\left\|y\right\|\le c.\tag1$$
However, in order to conclude we would need to show that $$\sup_{x\in\overline{B^X_1(0)}}\left\|Ax\right\|<\infty.\tag2 \label{eq2}$$
Why do we know that \eqref{eq2} cannot be infinity?
Because $A$ is linear, we have $$ \sup_{x \in \overline{B^X_1(0)}} \|Ax\| \leq \sup_{x \in B_2^X(0)} \|Ax\| = 2\cdot \sup_{x \in B_1^X(0)} \|Ax\|. $$