Question: Let E and F be two Banach spaces and let $T\in\mathcal{L}(E,F)$. Assume that $R(T)$ has finite codimension, i.e., there exists a finite-dimensional subspace $X$ of $F$ such that $X+R(T)=F$ and $X\cap R(T)=\{0\}$. Prove that $R(T)$ is closed.
Solution: Brezis' solution claims that:
Let $G=E\times X$ and consider the operator $$S(x,y)=Tx+y:G\longrightarrow F.$$ Applying the open mapping theorem, we know that $S$ is an open map, and thus $S(E\times (X\ \{0\}))=R(T)+(X\ \{0\})$ is open in $F$. Hence its complement, $R(T)$, is closed.
Some doubts I still have: First of all, what is the idea behind define the operator as the above? Moreover, I did not understand why $S$ is open and why this implies that $S(E\times (X\ \{0\}))=R(T)+(X\ \{0\})$ is open in $F$?
I appreciate any help. Thanks in advance.
Let me put my comments into a complete answer.
1) Idea behind defining the operator as above is to be able to write $F$ as a direct sum of 2 spaces.
2) Open mapping theorem states "If a continuous(bounded) map between Banach spaces is surjective then it is open." And notice $E\times X$ is Banach.
3) I think we should have $X/\{0\}$. Now because $E$ is the whole space it is both open and closed, again $X$ is both open and closed but $X/\{0\}$ is open because as you know $\{0\}$ is closed so it's complement in $X$ is open.