Let $K\colon X\to Y$ be a bounded linear operator between Banach spaces $X,Y$ and $\textrm{dim}(K(X))<\infty$.
It is well-known that $K$ is compact: If $U\subset X$ is bounded, then $\overline{K(U)}$ is compact in $K(X)$ since $K(X)\subset Y$ is a finite-dimensional subspace of $Y$.
I am wondering about the following:
Does this also imply that $\overline{K(U)}$ is compact in $Y$, i.e. that $K(U)$ is a relatively compact subset of $Y$? What happens if $K(X)$ is open in $Y$?
I could not find a satisfying answer yet.
Since we know that $K(U)$ is relatively compact as a subset of $K(X)$, it suffices to know the general topological fact that if $U \subset V \subset X$ where $X$ is a topological space and $U$ is compact in $V$ in the induced topology then $U$ is compact in $X$ also.
To this end, let $O := \{O_i: i \in I\}$ be a cover of $U$ consisting of open sets in $X$. Then $\bar{O} := \{\bar{O}_i := O_i \cap V: i \in I\}$ is an open cover of $U$ consisting of sets that are open in the induced topology on $V$. Since $U$ is compact in $V$, there exists a finite set $\bar{I} \subset I$ such that $\{\bar{O}_i : i \in \bar{I}\}$ is a finite open cover for $U$ in the topology of $V$. Then it is easy to see that $\{O_i: i \in \bar{I}\}$ is a finite subcover of $O$.