Denote $S$ by the set of normal states on $M_2(\Bbb C)$. Suppose $p(p\neq 0,1)$ is a projection in $M_2(\Bbb C)$ and $\rho\in S$. Define $S_p:=\{w\in S：w(p)=0\}$ and $d(\rho,S_p):=\inf_{\omega\in S_p}\|\rho-\omega\|$ .
It is east to see that $d(\rho,S_p)\geq 2\rho(p)$. Can $d(\rho,S_p)$ take any value in $[2\rho(p),2)$?
For any $c\in [2\rho(p),2)$, if we want to prove $d(\rho,S_p)=c$, we need to check that for any $\epsilon＞0$, there exists $\omega\in S_p$ such that $\|\rho–\omega\|＜c+\epsilon$. I was troubled by finding the suitable $\omega\in S_p$.
Because $M_2(\mathbb C)$ is finite-dimensional, all locally convex topologies agree on it. In particular, every state is normal. Every state is of the form $$ a\longmapsto \operatorname{Tr}(ha) $$ for some $h$ positive with $\operatorname{Tr}(h)=1$. If $\omega=\operatorname{Tr}(h\cdot)$ and $\omega(p)=0$, we have $$ 0=\operatorname{Tr}(hp)=\operatorname{Tr}(php). $$ As the trace is faithful, this means taht $php=0$, which in turn implies $hp=0$. It follows that $h=1-p$ (this is where we use that dimension is 2), and so $S_p$ is a singleton.
Given two states $\rho_j=\operatorname{Tr}(h_j\cdot)$, we have that $\|\rho_1-\rho_2\|=\operatorname{Tr}(|h_1-h_2|)$. So for general $\rho=\operatorname{Tr}(h\cdot)$ we have $$ \|\rho-\omega\|=\operatorname{Tr}(|1-p-h|). $$ Fix $\alpha\in[0,1]$ and consider $$ h=\alpha p+(1-\alpha)(1-p) $$ and $\rho=\operatorname{Tr}(h\cdot)$. Then $$ |1-p-h|=|-\alpha p+\alpha (1-p)|=\alpha\,1 $$ and $$ d(\rho,S_p)=\operatorname{Tr}(|1-p-h|)=2\alpha. $$