= Solution
The <holomorphic functional calculus> assigns to $a$ in a unital complex Banach algebra and every function holomorphic near $\sigma(a)$ the element
$$
f(a)=\frac1{2\pi i}\int_\Gamma f(z)(z1-a)^{-1}\,dz,
$$
where $\Gamma$ winds once around the spectrum. It is a unital algebra homomorphism, extends polynomial evaluation, is independent of the admissible contour, and obeys the <spectral mapping theorem>
$$
\sigma(f(a))=f(\sigma(a)).
$$
Let $z_K\in\mathcal R(K)$ be the coordinate function. For a character $\varphi$, put $\lambda=\varphi(z_K)$. If $\lambda\notin K$, then $(z_K-\lambda)^{-1}$ is one of the admitted rational functions, contradicting the fact that $z_K-\lambda$ lies in the kernel of $\varphi$. Thus $\lambda\in K$. For every rational function $r$ without poles on $K$,
$$
\varphi(r)=r(\lambda).
$$
Continuity of characters and uniform density extend this identity to every member of $\mathcal R(K)$. Conversely evaluation at every $\lambda\in K$ is a character. Therefore the <character space of R(K)> is naturally $K$.
<Runge approximation theorem> says that if $K\subset\mathbb C$ is compact, $f$ is holomorphic on a neighbourhood of $K$, and one chooses one point in every bounded component of $\mathbb C\setminus K$, then $f$ can be approximated uniformly on $K$ by rational functions whose finite poles belong only to the chosen points.
To prove it, surround $K$ by finitely many small rectangles contained in the domain of $f$ and apply the <Cauchy integral formula> on their oriented boundaries:
$$
f(z)=\frac1{2\pi i}\int_\Gamma\frac{f(\zeta)}{\zeta-z}\,d\zeta.
$$
Riemann sums approximate this integral uniformly on $K$ by rational functions with poles on $\Gamma$. If a pole $b$ lies in a component containing the selected point $a$, choose a polygonal path from $b$ to $a$ inside that component and subdivide it finely. The resolvent identity
$$
\frac1{z-b}-\frac1{z-c}
=\frac{b-c}{(z-b)(z-c)}
$$
allows the pole to be moved step by step along the path, with arbitrarily small uniform error on $K$. Moving every pole proves the theorem.
For an open set $U$, choose a compact exhaustion $K_1\subset K_2\subset\cdots$ such that every component of $\mathbb C\setminus K_n$ meets $\mathbb C\setminus U$. Runge's theorem approximates a holomorphic function on $K_n$ by a rational function with all poles outside $U$. A diagonal choice gives convergence uniformly on every compact subset, proving that such rational functions are dense in the <space of holomorphic functions> $\mathcal O(U)$ with its compact-open topology.
Finally, equality of the <Gelfand transform>[Gelfand transforms] gives
$$
\sigma_A(x_1)=\{\widehat{x_1}(\varphi):\varphi\in\Phi_A\}
=\{\widehat{x_2}(\varphi):\varphi\in\Phi_A\}
=\sigma_A(x_2).
$$
Let this common compact spectrum be $K$. The divided difference
$$
g(z,w)=
\begin{cases}
\dfrac{f(z)-f(w)}{z-w},&z\ne w,\\
f'(z),&z=w
\end{cases}
$$
is holomorphic near $K\times K$. The two-variable holomorphic functional calculus for the commuting pair $(x_1,x_2)$ gives an element $u=g(x_1,x_2)$ satisfying
$$
f(x_1)-f(x_2)=u(x_1-x_2).
$$
For every character,
$$
\widehat u(\varphi)
=g(\widehat{x_1}(\varphi),\widehat{x_2}(\varphi))
=f'(\widehat{x_1}(\varphi))\ne0.
$$
An element of a commutative unital Banach algebra is invertible exactly when its Gelfand transform has no zero. Thus $u$ is invertible, and $f(x_1)=f(x_2)$ implies $x_1=x_2$. This is <injectivity through a holomorphic functional calculus with nonvanishing derivative>.
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