Q:

Upper semi Fredholm operator is closed map

Let $T$ be a bounded linear operator from a Banach space X to X and given that $T(X)$ is closed and $ker(T)$ is finite dimensional.I want to prove that for any image of any closed set $M$ is closed i.e T takes closed sets to closed sets.

A:

The statement is not true.
Consider $\pi:\mathbb{R}^2\to \mathbb{R} = (x,y)\to y$  it is linear with finite dimensional kernel  and surjective.
Let $\Gamma \subset \mathbb{R}^2$ be the graph of the $\tan^{-1}(\cdot)$, then $\Gamma$ is closed  but $\pi(\Gamma)  = (-\frac \pi 2, \frac \pi 2)$ is not closed.
It is true though that if $S\lhd X$ is a closed subspace in $X$ then $T(S)$ is closed.
Indeed, since $Ker(T)$ is finite dimensional it is complemented in $X$ i.e. $X= Ker(T) \oplus K_1$ with $K_1\lhd X$ a closed subspace of X.
Notice that $K_1 \approx X/Ker(T)\approx T(X)$ (thanks to the open mapping theorem you gain that the obvious map is an homeomorphism).
Now, if $S$ is a closed subspace (hence a Banach) of $X$ then $T(S) = T(S + Ker(T))$ and $T(S+Ker(T))$ is closed since $\frac {S+Ker(T)} {Ker(T)}$ is a Banach space* hence complete hence closed as a subspace of $ \frac X{Ker(T)}\approx T(X)$
*If you quotient a Banach space by a closed subspace the result is still a  Banach.

