See 1.4.5 Theorem of Murphy's book : I want to prove that if $u$ be compact operator on $X$ which is Banach space and $\lambda\in \mathbb{C}\setminus\{0\}$, then $ker(u-\lambda)$ is finite-dimensional.
But in proof : if we assume that $Z=ker(u-\lambda)$, why we have $u(Z)\subseteq Z$ and $u_Z$ is in $K(Z)$ ?
Since $Z = \ker(u - \lambda)$ we have $(u-\lambda)z = 0$, so $u(z) = \lambda z \in Z$ for all $z \in Z$.
The restriction of a compact operator $u$ to a closed invariant subspace is compact: if $z_n$ is a bounded sequence in $Z$ then there is a subsequence such that $u(z_{n_k})$ converges in $X$. Since $Z$ is closed and $u(z_{n_k}) \in Z$, this subsequence converges also in $Z$. Thus, the restriction $u_Z$ of $u$ to $Z$ is compact, so $u_Z \in K(Z)$.
On the other hand $u_Z = \lambda \cdot \operatorname{id}_Z \in K(Z)$ and since $\lambda \neq 0$ we have that $\operatorname{id}_Z$ is compact, which is only possible in finite-dimensional spaces.