Referring to the Stone-Weierstrass theorem stated in this question.
From Rudin's functional analysis we have the following theorem:
5.7. Bishop Theorem Let $A$ be a closed subalgebra of $C(S)$. Suppose $g \in C(S)$ and $\left. g \right\rvert_E \subset A_E$ for every maximal $A$-antisymmetric set $E$. Then $g \in A$.
Here we have :
- $C(S)$ sup-normed Banach space of continuous complex functions on a compact Hausdorff space $S$
- $A \subset C(S)$ is an algebra if $fg \in A$ whenever $f, g \in A$.
- $E \subset S$ is $A$-antisymmetric if every $f \in A$ which is real on $E$ is constant on $E$
- If $A \subset C(S)$, $p, q \in S$ then $p \sim q$ if there's an $A$-antisymmetric set $E$ which contains $p$ and $q$. This defines a equivalence relationship and such equivalence classes are the *maximal $A$-antisymmetric sets.
I would assume the Stone Weierstrass theorem in the linked question is a special case of this one but I struggle to map the maximal $A$ antisymmetric part to that theorem however.
I'll take $C(S)$ to consist of real-valued functions. The case of complex-valued functions works similarly (note that you have an additional condition in the Stone-Weierstraß theorem, though).
Let $A_0\subset C(S)$ be a subalgebra that separates points and contains the constant functions, and let $A$ be the closure of $A_0$. Because scalar multiplication, addition and multiplication are continuous, $A$ is again an algebra.
Let $E\subset S$. If $x,y\in E$ with $x\neq y$, then there exists $f\in A$ such that $f(x)\neq f(y)$ since $A_0$ separates points. Thus $E$ cannot be $A$-antisymmetric. Hence every $A$-antisymmetric set is empty or a singleton. Since every singleton is clearly $A$-antisymmetric, the maximal $A$-antisymmetric sets are exactly the singletons.
As $A_0$ contains the constant functions, $A\vert_{\{x\}}$ contains all functions from $\{x\}\to\mathbb R$ for every $x\in S$. Thus $g\vert_E\in A\vert_E$ for every $g\in C(S)$ and every maximal $A$-antisymmetric set $E\subset S$. Therefore $A=C(S)$ by the Bishop theorem.