I have to prove this statement:
Let $X$ and $Y$ be banach spaces and let $T \in L(X, Y)$ be a linear continuous operator. If $dim (Y/T(X)) < \infty $, then $T(X)$ is closed in Y.
My ideas:
There is a more general fact that can be proved. Let $Y$ is a Banach space and $C,L$ are two linear subspaces in $Y$ that are algebraically complement to each other (i.e. $Y = C \oplus L$ in algebraic sence). Assume that $C$ is closed in $Y$ and $L$ is a Banach space in some norm $|| \cdot ||_L$ that dominates the norm in $Y$ (i.e. inclusion of $L$ into $Y$ is a continuous operator). Then $L$ is closed and $|| \cdot ||_L$ is equivalent to norm in $Y$. (Also I want to mention that this fact can be generalized to Frechet spaces).
Proof. Define $E = C \oplus L$ where norm on $C$ is induced from $Y$ and $L$ has norm $|| \cdot ||_L$. By definition $E$ is a Banach space and natural map $u:E \rightarrow Y$ that is induced by inclusions is continuous and bijective. Therefore $u^{-1}$ is continuous by closed graph theorem and it induces a linear homeomorphism between $L$ with induced norm from $Y$ and $L$ with norm $|| \cdot ||_L$. This ends the proof.
Now we can derive another fact from this. Let $X,Y$ be Banach spaces and $T:X \rightarrow Y$ be a continuous linear map s.t. $T(X)$ has a closed algebraic complement in $Y$. Then $T(X)$ is closed.
Proof. T(X) has a norm of quotient space $X / \ker T$ which dominates the norm of space $Y$ since $T$ is continuous. Therefore we can apply previous proposition.
Now it is easy to answer to your original question. Since $T(X)$ has finite codimension it has finite-dimensional algebraic complement. But all finite-dimensional subspaces are closed. And we are done.
