Proof, [Theorem 22.3, Zhu93]. Let $\seqj{x}\subset H$ be a maximal collection of non-zero vectors such that the spaces $\bracsn{Ax_j|j \in I}$ are mutually orthogonal. Such a collection exists by Zorn’s lemma, and must be at most countable given that $H$ is separable.

Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_{n}/2^{n}$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_{n} = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_{n} \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^{2}(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector.$\square$