If $X$ is a Banach space, $T \in \mathcal{B}(X)$, $T$ is compact, and $\lambda \neq 0$, then $T - \lambda I$ has closed range.
Proof with questions below
Proof: By (d) of Theorem 4.18, $\text{dim } \mathcal{N}(T - \lambda I) < \infty$. By (a) of Lemma 4.21, $X$ is the direct sum of $\mathcal{N}(T - \lambda I)$ and a closed subspace $M$. Define an operator $S \in \mathcal{M,X}$ by $$ Sx = Tx - \lambda x \;\;\; (1) $$ Then $S$ is one-to-one on $M$. Also, $\mathcal{R}(S) = \mathcal{R}(T - \lambda I )$. To show that $\mathcal{R}(S)$ is closed, it suffices to show the existence of an $r > 0$ such that $$ r \lVert x \rVert \leq \lVert S x \rVert \;\;\; (2) $$ by Theorem 1.26.
First bit I don't get
If (2) fails for every $r > 0$, there exists $\left\{ x_n \right\}$ in $M$ such that $\lVert x_n \rVert = 1$, $S x_n \to 0$, and (after a passage to a subsequence) $T x_n \to x_0$ for some $x_0 \in X$.
Why does such $x_n$ exist?
If (2) fails for every $r>0$, then we have $$\forall r>0, \ \exists x_r \in M \ \ni r \lvert x_r \rvert > \lvert Sx_r \rvert.$$ The $x_r$'s clearly can't be zero, so we can normalize them to have unit norm. Take a sequence $r_n \downarrow 0$ and let $x_n = x_{r_n}$. This should give everything except convergence of $Tx_n$ for a subsequence, which should follow from the compactness of $T$.