 Solution

Fix names $\dot{x},{\dot{a}}_1,\dots ,{\dot{a}}_n\in M$ and a first-order formula $\phi$. Using the syntactic forcing relation, form in $M$ the nameThis is a set by the axiom schema of separation in $M$. If $(\tau ,r)\in \dot{y}$ and $r\in G$, then the forcing theorem gives both ${\tau }^G\in {\dot{x}}^G$ and $\phi ({\tau }^G,{\dot{a}}_1^G,\dots ,{\dot{a}}_n^G)$. Conversely, if $z\in {\dot{x}}^G$ satisfies $\phi$, choose $(\tau ,q)\in \dot{x}$ with $q\in G$ and ${\tau }^G=z$. The truth direction of the forcing theorem supplies $s\in G$ forcing $\phi (\tau ,{\dot{a}}_1,\dots ,{\dot{a}}_n)$; directedness of $G$ gives $r\in G$ below both $q$ and $s$, so $(\tau ,r)\in \dot{y}$.

ThusEvery instance has such a witness, so separation in a generic extension proves $M[G]\models$ Separation.