Let $\mathbb{M} = (M, \in)$ be a class which models ZFC.
A forcing $\mathcal{P} = (P, \leq)$ on $M$ (where $P \in M$ and $\mathbb{M} \vDash P \in M$) is separative if for every $r \in P$ there are incomparable $p, q \geq r$. (I'm using the Shelah convention, which reverses the forcing relative to usual. Sorry!)
A subset $D \subseteq P$ is dense if for every $x \in P$ there is $d \in D$ with $x \leq d$.
A downward-closed directed set $G \subseteq P$ is $\mathcal{P}$-generic over $\mathbb{M}$ if it intersects every dense subset of $P$ which lies in $M$. (A set $G$ is directed if for any $p, q \in G$ we can find $r \geq p, r \geq q$.)
The lemma I am trying and failing to prove is:
If $\mathcal{P}$ is separative, then any $\mathcal{P}$-generic $G$ over $\mathbb{M}$ has $G \not \in M$.
My current working: let $G$ be a member of $M$. Then the set $P \setminus G$ exists in $M$ because $\mathbb{M}$ models ZF and so it lets us perform complementation.
Claim: $P \setminus G$ is dense.
Proof: given $p \in P$, we wish to find $q \geq p$ with $q \in P \setminus G$. If $p$ itself is not in $G$ then we're instantly done (let $q=p$). So wlog $p \in G \cap P$.
Now, since $\mathcal{P}$ is separative, we can find $x, y > p$ incomparable. We're done unless both of those are also in $G$, so suppose they are both in $G$.
$G$ is directed, so we can find $z > x, y$ with $z \in G$.
And that's as far as I've got; I haven't used downward-closedness, and I could assume for contradiction that $G$ intersects any given dense set if necessary. I'm quite sure this is easy.
You don't want the notion of incomparable, but incompatible (!). The elements $p$ and $q$, both in $P$, are incompatible if there is no $r \in P$ with $p \le r$ and $q \le r$. Incomparable is a weaker notion and just means that neither $p \le q$ nor $q \le p$ holds. That every element has incompatible elements above it is called the splitting condition, sometimes.
If $p \in P$ and $r_1, r_2 \ge p$ are the promised incompatible elements, then they cannot both be in $G$, as $G$ is directed (upwards) by definition. So at least one of them is in $P \setminus G$.