Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:
$A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.
I have proven this when $X$ is a reflexive space.
Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.
In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.
Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.
On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.
But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.
It's easily proved that if the image is compact, the operator is also compact.
But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?