= Solution
A unital \b[<Banach star-algebra>] is a complex unital <Banach algebra> with a continuous conjugate-linear <algebra involution> satisfying
$$
(ab)^*=b^*a^*,\qquad(a^*)^*=a,\qquad1^*=1.
$$
The algebra <norm> is complete and submultiplicative, with the unit conventionally normalized to <norm> one. Many definitions additionally require the involution to be isometric, $\|a^*\|=\|a\|$; the function algebra here satisfies this stronger convention as well. A <Banach star-algebra> does not by definition require the extra <C-star identity> $\|a^*a\|=\|a\|^2$.
Here is the lattice proof of the \b[<Stone-Weierstrass theorem>] in its complex self-adjoint form. Let $B$ be the uniform <closure> of the given unital star-subalgebra in $C(\Omega)$, and let $B_{\mathbb R}$ be its real-valued part. <Closure> preserves multiplication and <complex conjugation>, so $B$ is still a unital <star-algebra>. Real and imaginary parts lie in $B_{\mathbb R}$, and separation by the complex algebra implies separation by its real-valued part.
For $h\in B_{\mathbb R}$, <polynomial> approximation of absolute value on a bounded interval gives $|h|\in B_{\mathbb R}$. For example, this <polynomial> approximation follows from the binomial series for the square root, whose coefficients give <uniform convergence> on the relevant closed interval. Thus
$$
\max(h,k)=\tfrac12(h+k+|h-k|),\qquad
\min(h,k)=\tfrac12(h+k-|h-k|)
$$
also lie in $B_{\mathbb R}$. It is therefore a lattice of functions.
To see how the lattice gives density, fix a real continuous target $f$ and a tolerance $\epsilon>0$. For any two distinct points $x,y$, separation and constants provide $h_{xy}\in B_{\mathbb R}$ with $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$, by an affine rescaling of a separating function. For $x=y$, use the constant $f(x)$.
Fix $x$. The sets where $h_{xy}>f-\epsilon$ cover $\Omega$ as $y$ varies. A finite subcover and a pointwise maximum give $g_x\in B_{\mathbb R}$ with $g_x>f-\epsilon$ everywhere and $g_x(x)=f(x)$. The neighborhoods where $g_x<f+\epsilon$ now cover $\Omega$ as $x$ varies. A second finite subcover and a pointwise minimum give $h\in B_{\mathbb R}$ satisfying
$$
f-\epsilon<h<f+\epsilon\quad\text{everywhere on }\Omega.
$$
This is <lattice approximation from two-point approximation>. Using half the desired tolerance gives strict uniform error below the desired bound. Since $B_{\mathbb R}$ is closed, it contains every real <continuous function>. Splitting a complex function into real and imaginary parts proves $\boxed{B=C(\Omega)}$.
\b[The annulus test.] For the closed annulus, the family holomorphic in the interior is \b[not dense] in the full continuous-function algebra. A uniform limit on the annulus is locally uniform in its interior, so <locally uniform convergence of holomorphic functions> makes that limit holomorphic there. Thus the proposed family is closed in $C(\Omega)$. It is proper because $z\mapsto\overline z$ is continuous but not holomorphic on the annulus interior.
There is also an explicit continuous functional detecting the obstruction. Choose a small closed disk of radius $r$ contained in the annulus interior and let $L(h)$ be the integral of $h(z)\,dz$ around its positively oriented boundary. The <Cauchy integral theorem> gives $L(h)=0$ for every member of the holomorphic family, while parametrizing the circle gives $L(\overline z)=2\pi i r^2\ne0$. Since $|L(h)|\leq2\pi r\|h\|_\infty$, no <uniform approximation> of $\overline z$ by that family is possible.
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