= Solution
Write $K=\sigma_A(x)$ and $R(\lambda)=(\lambda1-x)^{-1}$. Choose a bounded finite union of small polygonal regions $V$ with $K\subset V$, $\overline V\subset U$ and $\partial V\cap K=\varnothing$. Its positively oriented boundary cycle $\Gamma$ has winding number one on $K$ and zero outside $V$; boundaries of holes have negative orientation. Such regions can be obtained from a sufficiently fine grid around the compact set $K$. This accommodates disconnected spectra and neighborhoods with holes.
Define the <holomorphic functional calculus> by the norm-convergent contour integral
$$
\Theta_x(f)=\frac1{2\pi i}\int_\Gamma f(\lambda)R(\lambda)\,d\lambda.
$$
The integrand is continuous on a finite rectifiable cycle, so the integral exists in the <Banach space> $A$. Its value is independent of the chosen cycle: on $U\setminus K$ the integrand is holomorphic, and cycles with the stated winding numbers are homologous there. Apply the scalar <Cauchy integral theorem> to every <bounded linear functional> and then use the <Hahn-Banach theorem> to recover equality of the algebra-valued integrals. Equivalently, a common refinement of the polygonal regions gives the same contour-deformation argument.
Linearity is immediate. For the <entire functions> $1,Z$, enlarge the integration cycle to a sufficiently large circle in the resolvent set, using their entire extension. Integrating the uniformly convergent <Neumann series> term by term gives
$$
\Theta_x(1)=1,\qquad \Theta_x(Z)=x.
$$
To prove multiplicativity, choose two nested admissible cycles, $\Gamma_o$ outside $\Gamma_i$, both surrounding $K$ and lying in $U$. The <resolvent identity> gives
$$
R(\lambda)R(\mu)=\frac{R(\mu)-R(\lambda)}{\lambda-\mu}.
$$
Use $\Gamma_o$ for $f$ and $\Gamma_i$ for $g$, multiply the two integrals, and insert this identity. In the term with $R(\mu)$, the outer <Cauchy integral formula> replaces $f(\lambda)/(\lambda-\mu)$ by $f(\mu)$. In the term with $R(\lambda)$, the inner integral of $g(\mu)/(\lambda-\mu)$ is zero because $\lambda$ is outside the inner region. Thus
$$
\Theta_x(f)\Theta_x(g)=\frac1{2\pi i}\int_{\Gamma_i}f(\mu)g(\mu)R(\mu)\,d\mu=\Theta_x(fg).
$$
The exchanges of integrals are justified by continuous integrands on compact separated cycles. This proves the unital algebra-homomorphism property, not just an identity of <algebra character> values.
For continuity in <local uniform convergence>, fix one cycle and bound
$$
\|\Theta_x(f)\|\leq\frac{\operatorname{length}\Gamma}{2\pi}\sup_\Gamma\|R(\lambda)\|\,\sup_\Gamma|f(\lambda)|.
$$
The last supremum is a defining compact-set <seminorm> of $\mathcal O(U)$. Hence the map is continuous.
If another continuous unital <algebra homomorphism> $\Psi$ sends $Z$ to $x$, its values on polynomials are forced. For every $\zeta\notin U$, the function $(Z-\zeta)^{-1}$ exists in $\mathcal O(U)$, so multiplicativity forces
$$
\Psi((Z-\zeta)^{-1})=(x-\zeta1)^{-1}.
$$
Therefore $\Psi$ and $\Theta_x$ agree on all rational functions with poles outside $U$. The permitted <Runge theorem> makes these functions dense in $\mathcal O(U)$ for <local uniform convergence>, including when $U$ is disconnected. Continuity makes the maps agree everywhere. Thus
$$
\boxed{\Theta_x\text{ is the unique continuous unital algebra homomorphism with }\Theta_x(Z)=x}.
$$
This proves the <continuity and uniqueness of holomorphic functional calculus> with the exact topology required.
For an <algebra character> $\phi$, multiplicativity and nonzeroness give $\phi(1)=1$. An invertible element cannot have zero <algebra character> value, so $\phi(x)\in\sigma_A(x)$. The same observation for each algebra element and the spectral bound from Question 3 gives $|\phi(b)|\leq\|b\|$; thus a <algebra character> is automatically a <bounded linear functional>. In particular,
$$
\phi(R(\lambda))=\frac1{\lambda-\phi(x)}.
$$
Apply $\phi$ inside the contour integral and use the <Cauchy integral formula> with winding number one at $\phi(x)$:
$$
\boxed{\phi(\Theta_x(f))=f(\phi(x))}.
$$
No separation of algebra elements by <algebra characters> is used; such separation can fail for a general commutative <Banach algebra>.
For the final clause, the two closed spectral pieces $X,Y$ are compact and have positive distance. Choose disjoint open neighborhoods $U_X,U_Y$ and let $f$ be identically one on $U_X$ and zero on $U_Y$. This locally constant function is holomorphic on their union, and $f^2=f$. Set $e=\Theta_a(f)$. Multiplicativity gives $e^2=e$, and the proved <algebra character> formula gives
$$
\boxed{\phi(e)=\begin{cases}1,&\phi(a)\in X,\\0,&\phi(a)\in Y.\end{cases}}
$$
This is the <spectral idempotent from a separated spectrum>. It is also nontrivial without assuming that <algebra characters> separate points. If $e=0$, choose $\lambda\in X$ and define $g=0$ on $U_X$, $g(z)=1/(z-\lambda)$ on $U_Y$. Then $(Z-\lambda)g=1-f$, so $(a-\lambda1)\Theta_a(g)=1$, with commuting factors, making $a-\lambda1$ invertible despite $\lambda\in\sigma_A(a)$. This is impossible. If $e=1$, exchange the two pieces and use $\lambda\in Y$ for the analogous contradiction. Hence $e\ne0,1$.
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