If $V$ and $W$ are Banach spaces, there is a pretty clean answer. Any bounded linear map $T: V \to W$ factors through a bounded linear map (same bound) $\tilde{T}: V/{\rm ker}(T) \to W$. Then $T$ will take closed subspaces to closed subspaces if and only if $\tilde{T}$ is bounded below, i.e., there exists $c > 0$ such that $\|\tilde{T}(x)\| \geq c\cdot \|x\|$ for all $x \in V/{\rm ker}(T)$.
The projection $\pi: V \to V/{\rm ker}(T)$ takes closed subspaces to closed subspaces, and if $\tilde{T}$ is bounded below then it also takes closed subspaces to closed subspaces, so $T = \tilde{T}\circ \pi$ must take closed subspaces to closed subspaces. That's one direction. Conversely, if $\tilde{T}$ is not bounded below then there exists a sequence $(x_n)$ in $V/{\rm ker}(T)$ with $\|x_n\| = 1$ for all $n$ and $\|\tilde{T}(x_n)\| \to 0$. Letting $V_0$ be the closed span of this sequence, $\tilde{T}$ takes $V_0$ bijectively onto its image, so by the Banach isomorphism theorem its image cannot be closed.
