Let $\mathfrak{A}$ be a [separable] unital C*-algebra and let $Q$ be a dense subset of the state space of $\mathfrak{A}$. Suppose that for each $f\in Q$ the associated GNS representation is faithful. Is $\mathfrak{A}$ simple?

I think the answer is negative.

If $A$ and $B$ are C*-algebras then the state space $S(A\oplus B)$ contains a natural copy of $S(A)$ and one of $S(B)$ such that $S(A\oplus B)$ is the convex hull of $S(A)\cup S(B)$. Moreover, the convex combinations $$ \sigma =\alpha \varphi +\beta \psi , $$ with $\varphi \in S(A)$, $\psi \in S(B)$, $0<\alpha ,\beta <1$, and $\alpha +\beta =1$, form a norm-dense subset of $S(A\oplus B)$.

The GNS representation of each such state $\sigma $ does not vanish on either $A$ or $B$ since neither does $\sigma $. In case both $A$ and $B$ are simple, then all of these representations will be faithful, providing a counter-example.

A very concrete example of this situation is simply $\mathbb C\oplus \mathbb C$!