I'm uncertain if my reasoning of this proof is correct. What do you think?
Problem
Let $X,Y$ be Banach spaces and $T: X \to Y$ a compact linear map. Suppose $Z\subset Y$ is a closed subspace such that $Z \subset T(X)$. Show that $\dim(Z) < \infty$.
Solution
Let $S$ be the inverse image of $Z$, i.e. $S = \{x \ |\ \exists z \in Z:T(x) = z \}$. Since $Z$ is a closed subspace, $S$ is a closed subspace as well. Hence, $S$ and $Z$ are Banach spaces. Now consider the map $M: S \to Z$ as the restriction of $T$. This map is surjective and compact (correct?).
Since $M$ is surjective (and bounded), we can apply the open map theorem. Hence $\exists r>0$ such that $B(0,r) \subset M(B(0,1))$. But $M(B(0,1))$ is relatively compact, hence $\overline{B(0,r)}$ is compact which means that $\overline{B(0,1)}$ is compact. Hence $Z$ must be finite dimensional.