I have an idea for a possible counterexample to the noncommutative Stone-Weierstrass problem. A good answer to the following question would really help.

Let $\mathcal{A}$ be the C-algebra of $2\times 2$ complex matrices, let $\mathcal{B}$ be the C-subalgebra of $2\times 2$ diagonal matrices, and let $v$ and $w$ be the unit vectors $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$, respectively. Then $f: A \mapsto \langle Av,v\rangle$ and $g: A \mapsto \langle Aw,w\rangle$ are pure states on $\mathcal{A}$. They are distinct, but $f(A) = g(A)$ for all $A \in \mathcal{B}$: we say that $\mathcal{B}$ fails to separate $f$ and $g$.

I would like to find a C-algebra $\mathcal{A}'$ which (unitally) contains $\mathcal{A}$ together with a C-subalgebra $\mathcal{B}'$ which contains $\mathcal{B}$, such that $\mathcal{B}'$ separates any extensions $f'$ and $g'$ of $f$ and $g$ to states on $\mathcal{A}'$. This could be trivially accomplished by setting $\mathcal{A}' = \mathcal{B}' = \mathcal{A}$, so let me add one more condition. Let $P = \left[\matrix{1&1\cr1&1}\right]$ and observe that the distance from $P$ to $\mathcal{B}$ is $1$. (Evaluation in the $(1,2)$ matrix entry is a norm one linear functional on $\mathcal{A}$ which takes the value $1$ on $P$ and is zero on $\mathcal{B}$.) To prevent trivial solutions to my problem, I impose the additional requirement that the distance from $P$ to $\mathcal{B}'$ must still be $1$.

Can we find a C-algebra $\mathcal{A}'$ which unitally contains $\mathcal{A}$ together with a C-subalgebra $\mathcal{B}'$ which contains $\mathcal{B}$, such that (1) the distance from $P$ to $\mathcal{B}'$ is $1$ and (2) $\mathcal{B}'$ separates any two states on $\mathcal{A}'$ which extend $f$ and $g$, respectively?