This question stems from 16.4 in Zhu's book on operator algebras.
Show that if $\mathcal{A}$ is a self-adjoint set of commuting operators in $B(\mathcal{H})$, then the weak-operator closure of $\mathcal{A}$ consists again of mutually commuting operators.
My idea is the following. take two elements $S$ and $T$ in the weak closure of $\mathcal{A}$. We can approx. both elements with $T_\alpha \overset{w}{\to} T$ and $S_\alpha \overset{w}{\to} S$
choose some element $T_\alpha$ of the sequence and $S$. We can approximate $S$ by some sequence $S_\beta$, so: $$\langle T_\alpha S x,y \rangle=\lim_{\beta}\langle T_\alpha S_\beta x,y\rangle$$ since all the elements of $\mathcal{A}$ commute (and each element of the sequence is an element of $\mathcal{A}$) then $$\lim_{\beta}\langle T_\alpha S_\beta x,y\rangle=\lim_{\beta}\langle S_\beta T_\alpha x,y\rangle=\langle S T_\alpha x,y\rangle$$
If we do the same for $T$ (writing it as a sequence of operators in $\mathcal{A}$ it will commute with any element $S_\alpha$) we get $\langle TS-STx,y\rangle=0$ which means $TS=ST$.
But this seems to prove things without using self-adjointness. If so, where is this proof wrong? In general, why do we need self-adjointness for which step of the proof?