I don't have my copy of Comfort & Negrepontis handy, so the following might be essentially the same as their proof, but I think it's clear enough.

The very rough idea is that, because $\mathcal F$ is a Q-point, the proof would be easy if each $A_i$ were a final segment of $\omega$ (Steps 3 and 4 below), and, because $\mathcal F$ is a P-point, we can reduce to this final-segment case by restricting to a suitable set in $\mathcal F$ (Steps 1 and 2 below).

Step 1. Get a big set $B$ that is almost included (i.e., included except for a finite set) in each $A_i$. This is essentially the fact that a selective ultrafilter is a P-point, plus the equivalence of a couple of definitions of P-point. In detail: Without loss of generality (because $\mathcal F$ is non-principal) $A_i$ has no elements $\leq i$, and so, for every natural number $x$, we can define $f(x)$ to be the smallest $i$ such that $x\notin A_i$. This $f$ cannot be constant on any set in $\mathcal F$, because the set on which it has value $i$ is disjoint from $A_i\in\mathcal F$. So $f$ is one-to-one on some $B\in\mathcal F$. On $B-A_i$, this one-to-one $f$ takes only values $\leq i$, so $|B-A_i|\leq i+1$. That completes step 1.

Step 2. Partition $\omega$ into a sequence of finite intervals $I_0,I_1,\dots$, where $I_n=[e_n,e_{n+1})$ (with $e_0=0$ and $e_{n+1}>e_n$), choosing the sequence $(e_n)$ growing so rapidly that, for all $i\leq e_n$, all elements of the finite set $B-A_i$ are $<e_{n+1}$. It is trivial to choose such $e_n$'s by induction on $n$. This choice ensures that, if $i\leq e_n$ and $j\geq e_{n+1}$ (thus, if $i<j$ are in $B$ and lie in $I$-intervals with at least one other $I$-interval between them) then $j\in A_i$.

Step 3. Apply selectivity to the function $g$ that is constant with value $n$ on exactly the interval $I_n$, to get a set $C\in\mathcal F$ that has at most one element in any $I_n$. So,if $i<j$ are in $B\cap C$ and are not consecutive in $C$, then $j\in A_i$.

Step 4. Finally, let $D$ consist of every second element of $C$, and let $E=C-D$. Because $\mathcal F$ is an ultrafilter, it contains exactly one of $D$ and $E$, so it contains one of $B\cap D$ and $B\cap E$. Within either of these sets, $i<j$ implies $j\in A_i$, so the proof is complete.