In his book Set Theory (second edition), page 203, Thomas Jech writes
Lemma 14.4 Let $\mathfrak{M}$ be a countable model of ZFC and $P$ be a partially ordered set. If $p\in P$, there exists a generic filter on $P$.
Then, the following exercise is given:
Exercise Let $\mathfrak{M}$ as above. Let $P\in M$ be a partially ordered set with no atom (for every $p\in P$, then there exist $q,r\in P$ such that $q,r\le p$ and $q,r$ are incompatible). If $G\subset P$ is generic over $\mathfrak{M}$, then $G\not\in \mathfrak{M}$.
The short proof of Lemma 14.4 is given here:
According to the exercise, the filter $G$ constructed in this proof must fail to belong to $\mathfrak{M}$ in some cases. But why ? Since $P\in \mathfrak{M}$ it seems that $G\in \mathfrak{M}$ using comprehension, doesn't it ?
If $G$ belongs to $\mathfrak{M}$, then so does $P \setminus G$. Since $P$ has no atoms, it is easy to show that $P \setminus G$ is dense in $P$, and therefore $G \cap ( P \setminus G ) \neq \emptyset$ by genericity, which is absurd!
The point is that in the proof of Lemma 14.4 you enumerated the dense subsets of $P$ which belong to $\mathfrak{M}$, and this enumeration will generally take place outside of $\mathfrak{M}$, and in the "real world". From this there is no reason to expect that the sequence of conditions of $P$ you base your generic set $G$ on belongs to $\mathfrak{M}$, and therefore there is little reason to expect that the generic set itself belongs to $\mathfrak{M}$. In fact, the exercise makes it clear that under fairly mild (and very common) assumptions the generic set will provably not belong to $\mathfrak{M}$.