I have tried to the following exercise :
Let $T$ be a compact operator, X an infite dimension Banach space and suppose that there is a closed subset $M$ such that $X= Im T\bigoplus M$, then $T$ is a finite rank operator.
Now my first approach to this was that since $M$ is closed we will have that $X/M$ will be a banach space and we know that $X/M \cong Im T$ but this is an algebraic isomorphism I am not sure we get an homeomorphism of spaces so that we could conclude that $Im T$ is a Banach space. If we have that $Im T$ we will have that this will be an open map from $X$ to $ImT$ and so $T(B_X(0,1))$ will be open in $Im T$ and so we have that for this to be relative compact $ImT$ has to have finite dimension. Now I am not sure that we will that $Im T$ will be a banach space because I think we get an algebraic isomorphism, will this be the case or do we have that $Im T$ is in fact a banach space? Thanks in advance.
The composition of a compact operator and a continuous operator is again compact.
Now consider $\pi \circ T$, where $\pi \colon X \to X/M$ is the canonical projection. By the above, it is compact, and by the assumption $X = \operatorname{Im} T \oplus M$ it is surjective. Also, $X/M$ is a Banach space, whence $\pi \circ T$ is open. Thus $X/M$ is locally compact, hence finite-dimensional.