Let $u: X \to X $ be a compact operator on a Banach space $X$ and let $\lambda \in \mathbb C$ be non zero. We know that $u-\lambda$ is Fredholm and that $X=\mathrm{ker}(u-\lambda)^n \oplus \mathrm{im}(u-\lambda)^n$ for some $n$. Consider the map $u\mid_{\mathrm{im}}=u\mid_{\mathrm{im}(u-\lambda)^n}$.
Why is the spectrum $\sigma(u\mid_{\mathrm{im}})$ closed in $\sigma (u)$?
Notice that $u|_{im}$ is compact because $u$ is compact.
If $u|_{im}x=\mu x$ for some $x\ne 0$, then $ux=\mu x$. Therefore, $\sigma(u|_{im})\subseteq \sigma(u)$.
If $ux=\mu x$ for some $\mu\in\mathbb{C}\setminus\{\lambda\}$ and some $x \ne 0$, then $x \in \mbox{im}$ because $$ x = \frac{1}{(\mu-\lambda)^{n}}(u-\lambda I)^{n}x. $$ Therefore, $\sigma(u)\setminus\{\lambda\} \subseteq \sigma(u|_{min})$.
Finally, notice that $u|_{im}$ does not have eigenvalue $\lambda$ because $u|_{im}-\lambda I$ is invertible on 'im'. So $\sigma(u|_{im})=\sigma(u)\setminus\{\lambda\}$, which is closed in $\sigma(u)$ because the non-zero eigenvalues of $u$ are isolated.