The problem is the next:
Suppose that $A\in B(X,Y)$, $K\in K(X,Y)$, where $X,Y$ are Banach spaces. Show that, if $A(X)\subset K(X)$, then $A\in K(X,Y)$.
Here $B(X,Y)$ stands for the set of all bounded linear operators from $X$ to $Y$, and $K(X,Y)$ stands for all compact operators from $X$ to $Y$.
I've tried some ways to prove it, but none of them successfully. What I reach to prove is the easy case, when $K$ is a surjective operator, because in this case$Y$ is finite-dimensional and every bounded linear operator on a finite-dimensional space is a compact operator. Then, I remember (but I'm not sure if it is true!) that a compact operator is one-to-one if and only if it is not surjective, so if we suppose that $K$ is not a surjective operator, then $K$ is one-to-one. I've tried to use this property but I could conclude nothing.
Any help or hint is well received.