I am trying to solve this problem from Gohberg's book:
Let $X$ be a Banach space and suppose that $T : \mathcal{D}(T) \subset X \to X$ has compact resolvent. Then, if $B$ is bounded on $X$, the operator $T+B$ is Fredholm with index $0$.
I tried the following approach: since $$(T-\lambda I)(T-\lambda I)^{-1} = I,~\forall \lambda \in \rho(T),$$ then, $$TR_{\lambda}(T) - \lambda R_{\lambda}(T) = I,$$ where $R_{\lambda}(T) := (T-\lambda I)^{-1}$.
Therefore, since $R_{\lambda}(T)$ is compact one has $$TR_{\lambda}(T) - I$$ compact. Furthermore, since $\lambda R_{\lambda}(T)$ is bounded and invertible, necessarily it is Fredholm with index zero.
Hence, $$TR_{\lambda}(T) - I$$ is Fredholm with index $0$, and it is also compact.
How to follow?
I figured out how to solve: in fact, it more or less standard to show that $B$ is $T$--compact. Furthermore, since $T$ has compact resolvent, therefore, for every $\lambda\in\rho(T)$, $T-\lambda I$ is Fredholm with index $0$. But since $\lambda I$ is $T$--compact, then $T$ is also Fredholm with index $0$. Analogously, the same holds for $T+B$. (see Theorem 4.3 on Gohberg's book, page 377).