In the book An Introduction to Operator Algebras by Kehe Zhu the following theorem is stated:
Theorem. Suppose $A$ is a closed subalgebra of a unital Banach algebra $B$ with $\hat{1}\in A$. If $x \in A$, then $\sigma_A(x)$ is the union of $\sigma_B(x)$ and some (possibly none) bounded connected components of $\mathbb{C}-\sigma_B(x)$; being $\sigma_A(x)$ the spectrum of $x$ in the Banach algebra $A$ and similarly with $\sigma_B(x)$.
The proof is based on the following fact (which is stated right before the theorem): Suppose $U,V \subset \mathbb{C}$ are open sets such that $U \subset V$ and $\partial U \cap V =\emptyset$. Then $U$ is the union of certain connected components of $V$.
Once said that, in the proof of the theorem it is proven that the open sets $U:=\sigma_A(x)^C$ and $V:=\sigma_B(x)^C$ satisfy the conditions of the above claim. Therefore, $U$ is the union of certain components of $V$.
Why does this imply the conclusion of the theorem?
First of all, if $\sigma_A(x)=\sigma_B(x)$ then we are done ( $\sigma_A(x)$ is the union of $\sigma_B(x)$ and none bounded components of $\sigma_B(x)^C$). Let's suppose that $\sigma_B(x) \subsetneqq \sigma_A(x)$, according to the statement, I have to show that if $V=\cup_{i\in I}V_i$, with $V_i$ the connected components of $V$ then $$ \sigma_A(x)=\sigma_B(x) \bigcup \left( \bigcup_{j\in J}{V_j} \right) $$ with $J\subset I$ and each $V_j$ bounded. I had the following idea: we can write $\sigma_A(x)=\sigma_B(x) \cup \left( \sigma_A(x)-\sigma_B(x) \right)$ and $\left( \sigma_A(x)-\sigma_B(x) \right) \subset V$. Now, let $V_i$ be a connected component of $V$ such that $V_i \cap \sigma_A(x) \neq \emptyset$. Then $V_i \cap \sigma_A(x)$ is a closed set in $V$, therefore if I could prove that $V_i \cap \sigma_A(x)$ is also an open set in $V$ then $V_i \subset V_i \cap \sigma_A(x)$ (since $V_i \cap \sigma_A(x)$ would be both: a closed set and an open set in $V$) and thus $V_i= V_i \cap \sigma_A(x)$; therefore we would have that the equation $\sigma_A(x)=\sigma_B(x) \cup \left( \cup_{j\in J}{V_j} \right) $ holds.
However, I don't see why $V_i \cap \sigma_A(x)$ would be an open set in $V$. Maybe my idea is just wrong.
Any suggestion in order to conclude the proof?
If I made any errors in my reasoning I would also appreciate any corrections.
In advance thank you.
I think we can complete the proof as follows:
Let $V_i$ a connected component of $V$ such that $\sigma_A(x) \cap V_i \neq \emptyset$. Then $$ \sigma_A(x) \cap V_i =V_i \cap \left( \sigma_A(x)-\sigma_B(x) \right) $$ As the user @RyszardSzwarc well said in the comments, we have that $\partial \sigma_A(x) \subset \sigma_B(x)$ and therefore $\sigma_A(x)-\sigma_B(x)=(\partial \sigma_A(x) \cup \sigma_A(x)^{\circ})-\sigma_B(x) =\sigma_A(x)^{\circ}-\sigma_B(x)=\sigma_A(x)^{\circ} \cap \sigma_B(x)^C$ which is an open set, thus we can write $\sigma_A(x) \cap V_i$ as an intersection of open sets. This completes the proof of the theorem.
Edit: It's probably good idea including to proof of $\partial \sigma_A(x) \subset \sigma_B(x)$. One way of proving this fact (probably not the optimal/easiest/elegant way) is as follows:
We can use the fact that $C$ is a unital Banach algebra and $(x_n) \subset C$ a sequence of invertible elements which converges to a noninvertible element $x \in C$, then $\lVert (x_n)^{-1} \rVert \rightarrow \infty$. Bearing this in mind,let $\lambda \in \partial \sigma_A(x)$, if $\lambda \notin \sigma_B(x)$ then $x-\lambda \hat{1}$ is an invertible element in $B$ but not in $A$. Since $\lambda \in \partial \sigma_A(x) $ we can find a sequence $(\lambda_n)$ such that $x-\lambda_n \hat{1}$ is an invertible element in $A$ but $x-\lambda_n \hat{1} \rightarrow x-\lambda \hat{1}$.
By the above claim, we would have that $\lVert \left( x-\lambda_n \hat{1} \right)^{-1} \rVert \rightarrow \infty$. On the other hand, if $G(B)$ is the set of invertible elements in $B$, then it's a known fact that the map $f:G(B) \longrightarrow G(B), y \mapsto y^{-1}$ is continuous and therefore $\lVert f(x-\lambda_n \hat{1}) \rVert \rightarrow \lVert f(x-\lambda \hat{1}) \rVert =\lVert \left( x-\lambda \hat{1} \right)^{-1}\rVert < \infty $. This condradiction shows that we must have $\partial \sigma_A(x) \subset \sigma_B(x)$.