If $\mathbb{P}\in M$ is atomless and the filter $G$ is $\mathbb{P}$-generic over M, then $G\not\in M$. [Kunen] If $\mathbb{P}\in M$ is atomless and the filter $G$ is $\mathbb{P}$-generic over M, then $G\not\in M$. 
So following is the proof by Kunen.
1.) $\mathbb{P}$ \ $G$ = $D$ is dense  (This can be easily shown by the fact that $\mathbb{P}$ is atomless.)
2.) Now if $G\in M$, then $D \in M$
3.) This contradict $G\ \cap\ D \neq \emptyset$
I don't quite understand (2), why if $G\in M$, then $D \in M$ ? Also if this where true where exactly is the contradiction in (3) ?
Thanks in advance,
Cheers.
 A: I suspect you're thinking too hard here. If $G\in M$, then we can apply Separation: the formula "$x\not\in G$" is a formula with a parameter from $M$, so by Separation in $M$ (and the fact that $\mathbb{P}\in M$) the set $$\{x\in \mathbb{P}: x\not\in G\}$$ is in $M$. But this set is exactly $D$.
Meanwhile, the contradiction in (3) is that $G$ is assumed to be $\mathbb{P}$-generic over $M$ - that is, for each $E\subseteq \mathbb{P}$ dense with $E\in M$, we have $G\cap E\not=\emptyset$. But - under the assumption $G\in M$ - the set $D$ is a dense subset of $\mathbb{P}$ in $M$ with $G\cap D=\emptyset$.

Going back to (1), note that the point is that if $G$ is in $M$, $M$ can define the set $D$; Separation is then used to show that $D$ actually exists in $M$. By contrast, if $G\not\in M$, how is $M$ supposed to define $D$? (Indeed, if $D$ is definable in $M$ then so is $G$, and by Separation if they are definable in $M$ then they are in $M$ since they're each subsets of a set in $M$.)
Another important note re: (1) is that in order to apply Separation we use the fact that $D$ is contained in a set in $M$ - namely, the poset $\mathbb{P}$. Of course in general, if $X\subseteq M$ is definable in $M$ we may not have $X\in M$ (e.g. take $X=M$).
