TITLE: Shelah's proof that proper forcing preserves P-points
QUESTION [5 upvotes]: In Proper and Improper forcing, VI.5; Claim 5.1 part 1 is the following:
If $F$ is a P-point in $V$, $P$ is a proper forcing notion and 
$\Vdash_P `` F$ generates an ultrafilter"
Then the ultrafilter generated by $F$ is a P-point in $V^P$. 
The proof starts like this: let $p\Vdash \{\tilde A_n : n<\omega\}$ be such that $p\Vdash \tilde A_n\in F$. 
By properness, for some $q\geq p$ and $A_{n,m}\in F$ for $n,m<\omega$ we have $q\Vdash \tilde A_n\in\{A_{n,m} : m <\omega\}$
To finish the proof, he uses that $F$ is a P-point in V, and finds an almost-subset of all the $A_{n,m}$. 
My question is how is the bolded statement true?  It appears (since each $A_{n,m}\in V$) that it would follow that P cannot add any subsets of $\omega$, which is clearly not true.  What am I missing?
EDIT: Thanks Mohammad Golshani for the comment.  I now believe that the bolded statement can be true.  But how do we get $q$ (and the collection of $A_{n,m}$)?  Another question is once we have $q$, why is $q$ in $G$?  Is the collection of such $q$ dense below $p$?

REPLY [5 votes]: In more details:  Let $N$ be a countable elementary submodel of some sufficiently large $H(\chi)$ containing $p$, $P$, $F$, and the sequence of names $\{\dot A_n:n<\omega\}$.  Then the countable set $N\cap F$ works for the collection of $A_{n, m}$, and any $(N,P)$-generic extension of $p$ will work for $q$.