subset of pure states with norm condition already dense I struggle to proof the following statement:
Let $Y\subseteq P\left(B\right)$ a subset of pure states on a $C^*$-Algebra $B$ such that for every $b\in B$ there exists a $\varphi \in P \left(B\right)$ such that $\left\Vert b\right\Vert = \varphi \left( b\right)$.
Claim: $Y\subseteq P\left(B\right)$ is already a dense subset.
 A: I'm assuming that the statement in your question is that "for every $b\in B^+$ there exists a $\varphi\in Y$ such that $\|b\|=\varphi(b)$". The positivity condition is necessary, because otherwise it may be impossible to have $\varphi(b)$ positive.
We can assume that $B$ is unital, since the state space of the unitization agrees with the state space of $B$. 
Suppose that $Y$ is not dense. Then the weak$^*$-closure $\overline Y$ is a proper weak$^*$-closed subset of $P(B)$. Moreover, $\text{conv}\,\overline Y$ is also proper because all elements of $P(B)$ are extreme. 
Let $\phi\in P(B)\setminus \text{conv}\,\overline Y$. By Hahn-Banach, there exists $c\in \mathbb R$ and $b\in B$ such that $\text{Re}\,\phi(b)>c>\text{Re}\,\psi(b)$ for all $\psi\in\text{conv}\,\overline Y$. We can rewrite this as 
$$
\phi(\text{Re}\,b) > c> \psi(\text{Re}\,b).
$$
By replacing $\text{Re}\,b$ with $\|b\|+\text{Re}\,b$, we may assume that $b\in B^+$ and that still
$$
\phi(b)>c>\psi(b),\ \ \ \psi\in\text{conv}\,\overline Y.
$$
Note that $\|b\|=\sup\{\phi(b):\ \psi\in P(B)\}$. By Banach-Alaoglu this sup is realized (choose a sequence with $\phi_n(b)\to \|b\|$ and use the compactness to get a convergent subsequence). So there exists $\phi_0\in \overline{P(B)}$ with $\|b\|=\phi_0(b)$. If $\phi_0\in \text{conv}\,\overline Y$ satisfies $\|b\|=\phi_0(b)$, then for any $\phi\in \overline{P(B)}\setminus \text{conv}\,\overline Y$ we would have
$$
\phi(b)>c>\phi_0(b)=\|b\|,
$$
a contradiction. It follows that, for every $\psi\in Y$ (actually, in the convex hull of its closure, but we don't need that now) there exists $c\in\mathbb R$ with
$$
\psi(b)<c<\phi_0(b)=\|b\|.
$$
So we have proven that $Y$ not dense implies that there exists $b\in B^+$ with $\psi(b)<\|b\|$ for all $\psi\in Y.$ Thus, if   for every $b\in B^+$ there exists $\varphi\in Y$ with $\varphi(b)=\|b\|$,  then $Y$ is dense in $P(B)$, 
