Atomless generic filters are not in the ground model

For any filter G on atomless P, $P\setminus G$ is dense. Given p, choose two incompatible refinements q,r of p. They cannot both be in G, since internal directedness would give a common stronger condition. At least one therefore lies in $P\setminus G$ below p. This is one finite existential argument for each p, not a simultaneous choice function.

If G were in M, internal Separation would make the actual set $D=P\setminus G$ an element of M. Step 1.1 makes D dense, while genericity would require $G\cap D\mathrm{\ne }\mathrm{\varnothing }$, impossible by its definition. Thus G is not in M. For singleton forcing, its unique filter P belongs to M and meets every dense set, exhibiting the failure when atomlessness is removed.