$A\in B(H)$ a unital abelian $C^*$-algebra with cyclic vector then $A'$ is abelian as well 
Let $A$ be a unital abelian $C^*$-subalgebra of $B(H)$ (with the same unit as
  that of $B(H)$), and assume there exists a vector $\xi \in H$ which is
  cyclic for $A$ (that is, $\{a\xi | a\in A \}$ is dense in $H$), then
  $A'$ is abelian as well.

First, I've proved a general lemma: Suppose $A$ is a unital $C^*$-algebra of $B(H)$ containing $1$. $\xi \in H$ is a cyclic vector for $A$ if and only if $\xi$ is separating for $A'$ (Here we don't need $A$ to be abelian).
Now, in order to show $\forall a,b\in A'$ , $ab=ba$, it suffices to show $ab\xi=ba\xi$ (as $\xi$ is cyclic for $A$ (unital) and therefore separating for $A'$.
My attempt: Let $\epsilon >0$. There exist $c,d\in A$ s.t. $||b\xi-c\xi||<\epsilon$ and $||d\xi-a\xi||<\epsilon$.
1) $||ab\xi-ac\xi||\leq ||a||||b\xi-c\xi||<\epsilon||a||$ 
2) $||ba\xi-bd\xi||\leq||b||||a\xi-d\xi||<\epsilon ||b||$ and 
3) $||ac\xi-bd\xi||\leq ||ac\xi-cd\xi||+||cd\xi-bd\xi||=||ca\xi-cd\xi||+||dc\xi-db\xi||\leq ||c||||a\xi-d\xi||+||d||||c\xi-b\xi||<||c||\epsilon+||d||\epsilon$.

Conclusion: if we could control the norms of $c,d$ then we're done.
However, I'm not sure how I can do that. Somehow my first instinct was to use somehow Kaplansly density theorem, but now I doubt that it is related.
Thank you in advance!
 A: (the argument below is extracted from Lemmas 7.2.13 and 7.2.14  of Kadison-Ringrose; the relevant more general theorems are Theorem 7.2.15 and Corollary 7.2.16)
If $x,y\in A'$ are selfadjoint, then we can find $\{a_n\}, \{b_n\}\subset A$, selfadjoint, with $a_n\xi\to x\xi$ and $b_n\xi\to y\xi$. Indeed, since $\xi$ is cyclic we can get $c_n$ in $A$ with $c_n\xi\to x\xi$. Then 
\begin{align}
\|c_n^*\xi-x\xi\|^2
&=\langle (c_n^*-x)\xi,(c_n^*-x)\xi\rangle
=\langle (c_n-x)(c_n^*-x)\xi,\xi\rangle\\ \ \\
&=\langle (c_n^*-x)(c_n-x)\xi,\xi\rangle
=\langle (c_n-x)\xi,(c_n-x)\xi\rangle\\ \ \\
&=\|c_n\xi-x\xi\|^2\to0
\end{align}
($c_n$ commutes with $c_n^*$ because $A$ is abelian, and with $x$ because $x\in A'$). Now we can take $a_n=(c_n+c_n^*)/2$ and do similar for $y$. 
Now, for any $c\in A$,
\begin{align}
\langle xy\,\xi,c\xi\rangle&=\langle y\xi,xc\xi\rangle
=\langle y\xi,cx\xi\rangle
=\lim_n\langle b_n\xi,ca_n\xi\rangle
=\lim_n\langle a_nb_n\xi,c\xi\rangle
=\lim_n\langle b_na_n\xi,c\xi\rangle\\ \ \\
&=\lim_n\langle a_n\xi,b_nc\xi\rangle
=\lim_n\langle a_n\xi,cb_n\xi\rangle
=\langle x\xi,cy\xi\rangle
=\langle x\xi,yc\xi\rangle
=\langle yx\xi,c\xi\rangle
\end{align}
(there is no issue with the joint limit since convergent sequences in a metric space are bounded). As $\{c\xi:\ c\in A\}$ is dense in $H$, we deduce that $xy\xi=yx\xi$. But $\xi$ is separating for $A'$, so $xy=yx$. As the selfadjoint elements of $A'$ span all of $A'$, it follows that $A'$ is abelian. 
As a remark, note that since $A$ is abelian, $A'\xi\supset A\xi,$ so $A'\xi$ is dense in $H$; that is, $\xi$ is cyclic for $A'$. So $\xi$ is separating for $A''$, and as $A\subset A''$, we have that $\xi$ is separating for $A$. That is, if $A$ is abelian and $\xi$ is cyclic for $A$, then it is also separating.
A: The only solution I can see uses Tomita-Takesaki theory: Let $M=A''$ be the von Neumann algebra generated by $A$, so that $M$ is also abelian, i.e., $M\subseteq M'$.
Note that $M'=A'''=A'$, so we are done if we show that $M'=M$ (in fact, this implies that $M$ is maximal abelian). We already have one inclusion $M\subseteq M'$.
Since $\xi$ is cyclic for $A$, it is also cyclic for $M$. Since $\xi$ is separating for $A'$, then it is separating for $M\subseteq A'$.
Let $J:H\to H$ be the modular conjugation of Tomita-Takesaki, i.e., $J=J^{-1}$ and $M'=JMJ$, so that
$$M'=JMJ\subseteq JM'J=JJMJJ=M$$
therefore $M=M'$ is abelian.

In fact, these arguments prove the following: If $M$ is an abelian von Neumann algebra of $B(H)$ with a cyclic vector, then $M=M'$,i.e., $M$ is maximal abelian.
