Let $\xi,\eta\in\mathcal H$ where $\eta\perp\overline{A'\xi}$ where $A\subset\mathcal B(\mathcal H)$ is a unital $*$-algebra. How do I show that there is a sequence $\{ a_n\}\subset A_{sa}\cap(A)_1$ such that $\|\xi-a_n\xi\|\leq n^{-1}$ and $\|a_n\eta\|\leq n^{-1}2^{-n}$ using the Kaplansky's density theorem?
Here $A_{sa}$ is the self-adjoint operators in $A$ and $(A)_1$ is the closed unit ball of $A$.
I can see that $[A'\xi]\in (A'')_1\cap (A'')_{sa}=\overline{(A)_1}^{SOT}\cap \overline{A_{sa}}^{SOT}$ using the theorem. But I don't see how I can extract the sequence from $A_{sa}\cap(A)_1$ satisfying the above conditions.
The key observation is that $A_{sa}$ is convex, so its sot closure agrees with its wot. So, if $p$ is your projection, there exists a net $\{b_j\}$ such thaht $$b_j\to p\ \ \text{ sot};$$ then $b_j\to p$ wot, and so $b_j^*\to p$ wot. Then $$(b_j+b_j^*)/2\to p$$ gives a net of selfadjoints that converges wot to $p$. As $\overline{A_{sa}}_{\rm sot}=\overline{A_{sa}}^{\rm wot}$, there exists a net $\{a_j'\}$ in $A_{sa}$ with $a_j\to p$. Now applying the proof of Kaplansky to $a_j'$, we let $$ a_j=f(a_j'), $$ where $$ f(t)=\begin{cases} 1,&\ t\geq1\\ t,&\ -1\leq t\leq 1\\ -1,&\ t\leq -1 \end{cases} $$ Then we have a new net $\{a_j\}$ with $\|a_j\|\leq1$ and still $a_j\to p$, since $$ f(a_j)\to f(p)=p. $$
Finally, when you specialize $\xi$, you can extract a subsequence $\{a_n\}$ of $\{a_j\}$ such that $a_n\xi\to p\xi$.