It's a very plausible conjecture, but I think it's false. Consider the C${}^\ast$-algebra $K(l^2)$ of compact operators on $l^2$. For any unit vector $v \in l^2$, the map $A \mapsto \langle Av,v\rangle$ is a pure state. An arbitrary state has the form $A\mapsto {\rm Tr}(AB)$ for some positive trace-class operator $B$ with unit trace. For any vector $v$ let ``$v\otimes v$'' denote the trace-class operator $w \mapsto \langle w,v\rangle v$, so that ${\rm Tr}(A\cdot v\otimes v) = \langle Av,v\rangle$.
Fix $n$ and consider the trace-class operators $B$ with $\langle Be_j,e_k\rangle = 0$ for all $j$ and $k$ with either $j > n$ or $k > n$. This set can be identified with the space of $n\times n$ complex matrices, which has complex dimension $n^2$. If $v$ is any of the $n^2$ unit vectors $e_j$ for $1 \leq j \leq n$, $\frac{1}{\sqrt{2}}(e_j + e_k)$ for $1 \leq j < k \leq n$, $\frac{1}{\sqrt{2}}(e_k + ie_j)$ for $1 \leq j < k \leq n$ then $v\otimes v$ is a positive, unit trace, trace-class operator, and these $n^2$ operators span the set of all $n\times n$ matrices.
Let $v_1$, $\ldots$, $v_{n^2}$ be those $n^2$ unit vectors, making sure to set $v_1 = e_1$. Then for $1 \leq j \leq n^2$ let $w_j = \alpha(v_1 + \frac{1}{n}v_j)$, where $\alpha$ is chosen so that $w_j$ will have unit norm. This gives us $n^2$ unit vectors, all very close to $e_1$, whose corresponding trace-class operators still span the full $n\times n$ matrix algebra.
Do the above for all $n$ and then let $C$ be the set of all the $w$ vectors ($n^2$ of them for each value of $n$). Then $C' =$ the set of corresponding pure states is compact with the single cluster point $e_1\otimes e_1$. But the linear span of $C'$ contains the full $n\times n$ matrix algebra for all $n$, so in particular it contains the trace-class operators $e_n\otimes e_n$ for all $n$, an infinite discrete sequence which converges weak${}^\ast$ to $0$. So any face of the state space that contains $C'$ will contain $e_n \otimes e_n$ for all $n$, but not $0$, so it can't be compact.
