This is the exercise:
Suppose $x\in A$, $x_n\in A$, and $\lim x_n=x$. Suppose $\Omega$ is an open set in $\mathbb C$ that contains a component of $\sigma(x)$. Prove that $\sigma(x_n)$ intersects $\Omega$ for all sufficiently large $n$. (This strengthens Theorem 10.20.)
Hint: If $\sigma(x)\subset\Omega\cup\Omega_0$, where $\Omega_0$ is an open set disjoint from $\Omega$, consider the function $f$ that is $1$ in $\Omega$, and $0$ in $\Omega_0$.
According to Rudin's hint, we can define $f\in H(\Omega\cup\Omega_0)$ as follows: $$ f(\lambda) = \left \{\begin{array}{lll} 1 & \ \ \ \ \lambda\in \Omega\\ 0 & \ \ \ \ \lambda\in \Omega_0\\ \end{array}\right. $$ And then by symbolic calculus we have $$ \tilde{f}(x) = \left \{\begin{array}{lll} e & \ \ \ \ \sigma(x)\subset \Omega\\ 0 & \ \ \ \ \sigma(x)\subset \Omega_0\\ \end{array}\right. $$ For the $x\in A$ that $x_n\rightarrow x$, we have $$\tilde{f}(x)=\tilde{f}(\lim_{n\to\infty}x_n)=\lim_{n\to\infty}\tilde{f}(x_n),$$ meaning $\{\tilde{f}(x_n)\}$ as a sequence in $A$ converges to $\tilde{f}(x)$. So for a large enough $n$; $$\sigma\big(\tilde{f}(x_n)\big)\cap\sigma\big(\tilde{f}(x)\big)\neq\emptyset.$$ And by spectral mapping theorem; $$f\big(\sigma(x_n)\big)\cap f\big(\sigma(x)\big)\neq\emptyset,$$ $$f\big(\sigma(x_n)\big)\cap \{0,1\}\neq\emptyset.$$ So we are forced to have only one of these three possibilities;
1) $f\big(\sigma(x_n)\big)=\{0,1\}$
2) $f\big(\sigma(x_n)\big)=\{1\}$
3) $f\big(\sigma(x_n)\big)=\{0\}$
The third case never happens. Because if it happens by the definition of $f$, we can conclude for any large $n$ that $$\sigma(x_n)\subset\Omega_0,$$ in which by tending $n$ to infinity, we get the following contradiction; $$\sigma(x)\subset\Omega_0.$$ In the first and second cases, there will be at least one $\lambda\in \sigma(x_n)$ such that $f(\lambda)=1$ which by definition simply means $\sigma(x_n)$ intersects $\Omega$.
Is this true? Thank you for your help.