I'm trying to understand the proof of Lemma 4.4 in Solovay's famous paper "A Model of Set-Theory in Which Every Set of Reals is Lebesgue Measurable". The lemma is slightly more complicated, but it basically says that if $M[G]$ is some generic extension (with $G\subseteq P_1$), and $M[s]$ is a subextension, then there is a subset $\Sigma\subseteq P_1$ (in $M[s]$) s.t $G$ is generic with respect to it (and hence $M[G]=M[s][G]$ is a generic extension of $M[s]$).
To build $\Sigma$, he builds some increasing sets of "bad element" to remove from $P_1$.
I don't understand why we need more than 1 step. If we take an union of dense subsets over each element of a dense subset, we get a dense subset, right? So all of the elements should be in $A_1$.
More formally, suppose that $p\in A_2$. Then by definition there is a dense subset $X$ s.t for every $q\ge p$ in $X$, $q\in A_1$. but for each $q\in A_1$ there is a dense subset $Y_q$ s.t. for every $r\ge q$ in $Y_q$, $r\in A_0$.
Hence, if we take the dense subset $D=\left\{r: \ if\ r\ge p\ then\ there\ is\ q\in X\ s.t\ r\ge q\ge p\ and\ r\in Y_q\right\}$ , we can get that $p\in A_1$.
What am I missing?