Fredholm and Compact Operators

Question

Let $X$ and $Y$ be Banach spaces and $T\in B(X,Y)$ be Fredholm. Then there is $S\in B(Y,X)$ such that $ST=I+K_{1}$ and $TS=I+K_{2}$ where $K_{1},K_{2}$ are compact operators.

Proof: Since $T$ is Fredholm then $KerT$ and $Coker(ImT)$ are finite diminsional and hence complimented, so assume $X=KerT\oplus V$ and $Y=ImT\oplus W$. Now define $T_{v}$ to be the restriction of the map $T$ to $V$ that is $$T_{v}:V\longrightarrow ImT$$ by definition we have $T$ is a bijection and $V$ and $ImT$ are Banach spaces then by the open mapping theorem $S_{1}=T_{v}^{-1}$ is bounded, Now consider $S:Y\longrightarrow X$ defined by $$S(y)=S_{1}P(y)$$ where P is the projection from $Y\longrightarrow ImT$ QED.

Question: a) If $T$ is Fredholm and $Ind(T)=n(T)-d(T)=0$, how can I change the proof to say that $S$ can be chosen to be invertible not only bounded. where $n(T)$ is the dimention of the kernel and $d(T)$ is the dimension of the cokerT.

b) Let $X$ be a Banach space and $T\in B(X)$ show that$$\bigcap_{K\in K(X)}\sigma(T+K)=\sigma(T)-\{\lambda:\lambda I-T\text{ is Fredholm of Index zero}\}.$$


Note: I'm really stuck on part a, as for part b) I'm trying to use the fact that $T$ is Fredholm if and only if $T+K(X)$ is invertible in $B(X)/K(X)$ I know that I'm very close but not there yet.