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If $V$ and $W$ are Banach spaces, I can say the following. If $T(V)$ is finite-dimensional then every subspace of $T(V)$ is closed, so $T$ does take all closed subspaces to closed subspaces. If ${\rm ker}(T)$ is finite-dimensional, factor $T$ through a bounded linear map $\tilde{T}: V/{\rm ker}(T) \to W$; 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 $x \in V/{\rm ker}(T)$.

Proof: Suppose $\tilde{T}$ is bounded below and let $V_0$ be any closed subspace of $V$. Since ${\rm ker}(T)$ is finite-dimensional, $V_0 + {\rm ker}(T)$ is closed and so the projection $\pi: V \to V/{\rm ker}(T)$ takes $V_0$ to a closed subspace of the quotient, and since $\tilde{T}$ is bounded below it also takes closed subspaces to closed subspaces, so $T = \tilde{T}\circ \pi$ must take $V_0$ to a closed subspace. 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. QED

If neither ${\rm ker}(T)$ nor $T(V)$ is finite-dimensional, I think there will have to be a closed subspace of $V$ whose image is not closed, but I need to think about that a little more.

If $V$ and $W$ are Banach spaces, the complete answer goes as follows. (1) If $T(V)$ is finite-dimensional then every subspace of $T(V)$ is closed, so $T$ does take all closed subspaces to closed subspaces. (2) If ${\rm ker}(T)$ is finite-dimensional, factor $T$ through a bounded linear map $\tilde{T}: V/{\rm ker}(T) \to W$; 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 $x \in V/{\rm ker}(T)$. (3) If neither $T(V)$ nor ${\rm ker}(T)$ is finite-dimensional, then there is always a closed subspace of $V$ whose image under $T$ is not closed.

Proof of (2): Suppose $\tilde{T}$ is bounded below and let $V_0$ be any closed subspace of $V$. Since ${\rm ker}(T)$ is finite-dimensional, $V_0 + {\rm ker}(T)$ is closed and so the projection $\pi: V \to V/{\rm ker}(T)$ takes $V_0$ to a closed subspace of the quotient, and since $\tilde{T}$ is bounded below it also takes closed subspaces to closed subspaces, so $T = \tilde{T}\circ \pi$ must take $V_0$ to a closed subspace. 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. QED

Proof of (3): according to the answer to this question, if neither $T(V)$ nor ${\rm ker}(T)$ is finite-dimensional, then there exists a closed subspace $V_0$ of $V$ such that $V_0 + {\rm ker}(T)$ is not closed. Then $T^{-1}(T(V_0)) = V_0 + {\rm ker}(T)$ is not closed, which shows that $T(V_0)$ cannot be closed. QED

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.

If $V$ and $W$ are Banach spaces, I can say the following. If $T(V)$ is finite-dimensional then every subspace of $T(V)$ is closed, so $T$ does take all closed subspaces to closed subspaces. If ${\rm ker}(T)$ is finite-dimensional, factor $T$ through a bounded linear map $\tilde{T}: V/{\rm ker}(T) \to W$; 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 $x \in V/{\rm ker}(T)$.

Proof: Suppose $\tilde{T}$ is bounded below and let $V_0$ be any closed subspace of $V$. Since ${\rm ker}(T)$ is finite-dimensional, $V_0 + {\rm ker}(T)$ is closed and so the projection $\pi: V \to V/{\rm ker}(T)$ takes $V_0$ to a closed subspace of the quotient, and since $\tilde{T}$ is bounded below it also takes closed subspaces to closed subspaces, so $T = \tilde{T}\circ \pi$ must take $V_0$ to a closed subspace. 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. QED

If neither ${\rm ker}(T)$ nor $T(V)$ is finite-dimensional, I think there will have to be a closed subspace of $V$ whose image is not closed, but I need to think about that a little more.

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 $\pi: V \to {\rm ker}(T)$ 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.

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.