An operator has closed range if and only if the image of some closed subspace of finite codimension is closed.

Let $B$ be a Banach space, $H,K$ be closed subspaces and let $K$ be finite dimensional.

Suppose $B = H\oplus K$ and $T:B\to B$ is a bounded linear operator.

How do I show that $T(B)$ is closed $\iff$ $T(H)$ is closed ?

The internal sum of a closed subspace and a finite-dimensional subspace is closed, so if $T(H)$ is closed then $T(B) = T(H) + T(K)$ is closed. (See the solution to exercise 41 here or the comments below for a proof).

For the other direction, we can assume that $T$ is injective, otherwise we consider the factorization of $T$ over $B/\ker T = (H / (H \cap \ker T)) \oplus (K / (K \cap \ker T))$ (noting that $T$ and its factorization have the same image).

Suppose that $T(B)$ is closed. Then $T(H)$ is a subspace of $T(B)$ of finite codimension, so it has an algebraic complement $Z$, so $T(B) = T(H) \oplus Z$ as vector spaces. Since $Z$ is finite-dimensional, $Z$ is closed, so the linear operator $S \colon H \oplus Z \to T(B)$ defined by $S(h,z) = T(h) + z$ is a continuous linear bijection between Banach spaces, hence it is a homeomorphism by the open mapping theorem. Since $H$ is closed in $H \oplus Z$, we have that $T(H) = S(H,0)$ is closed in $T(B)$.