If $A$ is a unital Banach algebra and $B$ is a closed subalgebra and $\sigma$ denotes the spectra then the following inclusion holds:
$$ \partial \sigma_B (b) \subseteq \partial \sigma_A (b)$$
for every $b \in B$. Consider the following proof of the statement:
where condition (1) and (2) are
(1) The set $\mathrm{Inv}(B)$ is a clopen subset of $B \cap \mathrm{Inv}(A)$
(2) For each $b \in B$: $ \sigma_A (b) \subseteq \sigma_B (b) $ and $ \partial \sigma_B (b) \subseteq \partial \sigma_A (b)$
Where in the proof is it used that $b-\lambda \notin \mathrm{Inv}(A)$? It holds that $\mathrm{Inv}(B) \subseteq \mathrm{Inv}(A)$ so that if $b-\lambda_n \in \mathrm{Inv}(B)$ then also $b-\lambda_n \in \mathrm{Inv}(A)$ and hence $\lambda \in \partial \sigma_A(b)$ since $(b-\lambda_n) \to (b-\lambda)$.
The first thing in the proof concerning $A$ is $b - \lambda \notin \operatorname{Inv}(A)$, which just means $\lambda \in \sigma_A(b)$.
Next, it is observed that the terms of the sequence $(\lambda_n)$ do not belong to $\sigma_A(b)$ - namely, it is stated that $b - \lambda_n \in \operatorname{Inv}(A)$ - which says that $\lambda$ is not an interior point of $\sigma_A(b)$. Since $\lambda \in \sigma_A(b)\setminus \sigma_A(b)^{\Large\circ}$, it follows that $\lambda \in \partial \sigma_A(b)$.
$b-\lambda \notin \operatorname{Inv}(A)$, or equivalently $\lambda\in\sigma_A(b)$ is essential for the argument. If we had $b-\lambda\in \operatorname{Inv}(A)$, or, equivalently, $\lambda\notin\sigma_A(b)$, then we could not have $\lambda\in\partial\sigma_A(b)$, since spectra are closed.