Compact operator with closed range has finite dimensional range

Let $X,Y$ be Banach Spaces, and let $T\in K(X,Y)$ be a compact operator from $X$ to $Y$. I have to prove that $T(X)$ is closed in Y if, and only if, $\dim(T(X))<\infty$.

Can anybody help me with this proof, please? There is surely some property I haven't thought about, but I'm getting really weird right now... Thank you!

Let $Z=T(X)$. Then $Z$ is also a Banach space, as a closed subspace of a Banach space, and $T:X\to Z$ is onto, and hence open, due to Open Mapping Theorem. If $Z$ were infinite dimensional, then $T$ would not be compact, as open sets in infinite dimensional spaces are not pre-compact.

Hint: Try the open mapping theorem.

Hint: an infinite dimensional Banach space is never $\sigma$-compact (by Baire category theorem).