= Solution
\b[False as printed: separativity alone does not guarantee a new generic filter.] A <separative forcing order> means that if $p\not\leq q$, some $r\leq p$ is incompatible with $q$. The one-point order $\mathbb P=\{p\}$ is separative vacuously. Its only <generic filter> is $\{p\}=\mathbb P$, and this belongs to $M$. Thus no $G\notin M$ exists in this example.
The natural missing hypothesis is that the nonempty order is an <atomless forcing order>. Under that hypothesis the intended argument works. Since $M$ is a <countable transitive model>, enumerate externally its dense subsets as $D_0,D_1,\ldots$. Choose $p_{n+1}\leq p_n$ with $p_{n+1}\in D_n$. Then
$$
G=\{q\in\mathbb P:\exists n\ p_n\leq q\}
$$
is a directed upward-closed <filter in an ordered set> and meets every $D_n$. This is the <Rasiowa–Sikorski lemma>, giving a <generic filter> over $M$.
If $G\in M$, the <set difference> $D=\mathbb P\setminus G$ would belong to $M$. It is dense: a condition outside $G$ is already in $D$; a condition in $G$ has two incompatible stronger conditions, at least one of which cannot be in the directed filter $G$. Genericity would require $G\cap D\ne\varnothing$, a contradiction. Thus
$$
\boxed{\text{with atomlessness, a generic }G\text{ exists and }G\notin M.}
$$
This uses <generic filter for an atomless order is new>. A <forcing atom> is precisely the obstruction illustrated by the counterexample; separativity does not remove it.
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