TITLE: $T(X)$ has a closed linear complement in $Y$ then $T(X)$ is closed in $Y$
QUESTION [1 upvotes]: I am trying to understand the proof of the following claim:

Let $X$ and $Y$ be Banach spaces and the linear operator $T : X \rightarrow Y$ be continuous. If $T(X)$ has a closed linear complement in $Y$, then $T(X)$ is closed in $Y$. In particular, if $T(X)$ has finite codimension in $Y$, then $T(X)$ is closed in $Y$.

The proof is attached below. I don't understand they introduce the operator $S$. Can't they just start the proof from highlighted 3 lines?

Theorem 8 is:

REPLY [1 votes]: They seem to be introducing $S$ (which is surjective and therefore has closed image) in order to apply Theorem 8 to obtain $M$.  We can't start at the highlighted portion because we can't be sure whether such an $M$ exists unless we know that $T(X)$ is closed, which is what we're trying to prove.
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