Is the following claim and proof correct?? I can't seem to find an error.
Let $X$ be a complex Banach space, and $A\in \mathcal{B}(X)$ a bounded operator. Then, if $\lambda\in \sigma(A)$ is isolated in the spectrum, $\lambda \in \sigma_{ap}(A)$.
Attempted proof. Since $\lambda$ is isolated in $\sigma(A)$, there exists $r>0$ such that function $z\mapsto A_z:=(A-z)^{-1}$ is holomorphic on the punctured open disk $\Delta_r:=\{z\in \mathbb{C}: 0<|z-\lambda|<r\}$.
Assume $\lambda\notin \sigma_{ap}(A)$. That is $A_\lambda:=A-\lambda$ is not surjective but is bounded below (meaning injective and has closed range). That is $$\delta_\lambda:=\inf_{\|x\|=1}\|(A-\lambda)(x)\|>0.$$ Then, for every sequence $(z_n)_{n\in\mathbb{N}}\subset \Delta_r$ converging to $\lambda$, such that $|z_n-\lambda|< \frac{\delta_\lambda}{2}$, and for every $x\in X$ with $\|x\|=1$, $$0<\delta_\lambda-\frac{\delta_\lambda}{2}<\|(A-\lambda)(x)\|-|z_n-\lambda|\leq \|(A-z_n)(x)\|$$ Hence, for all $n\in\mathbb{N}$, $\inf_{\|x\|=1} \|(A-z_n)(x)\|>\frac{\delta_\lambda}{2}$, and so $$\|(A-z_n)^{-1}\|=\sup_{x\neq 0}\frac{\|x\|}{\|(A-z_n)(x)\|}=\sup_{\|x\|=1}\frac{1}{\|(A-z_n)(x)\|}=\frac{1}{\inf_{\|x\|=1} \|(A-z_n)(x)\|}<\frac{2}{\delta_\lambda}$$
As a result, $(A-z_n)^{-1}$ is a Cauchy sequence, since $$\|(A-z_n)^{-1}-(A-z_m)^{-1}\|\leq |z_n-z_m|\|(A-z_n)^{-1}\|\|(A-z_n)^{-1}\|<|z_n-z_m|\left(\frac{2}{\delta_\lambda}\right)^2.$$
Then taking limit in the identity $$(A-z_n)(A-z_n)^{-1}=(A-z_n)^{-1}(A-z_n)=Id_X,$$ and, due to continuity, we conclude $A-\lambda$ is invertible.
Contradiction. $\square$