= Solution
Use the isolating sets from the preceding part and make the finite disjointification
$$
B_n=A_n\setminus\bigcup_{j<n}A_j.
$$
For $j<n$, $A_j\notin U_n$, so $A_j^c\in U_n$. Therefore
$$
B_n=A_n\cap\bigcap_{j<n}A_j^c\in U_n
$$
by finite-intersection closure of the <ultrafilter>. If $m\ne n$, $B_n\subseteq A_n$ and $A_n\notin U_m$ imply $B_n\notin U_m$. Also $B_n$ avoids every earlier $A_j$, hence every earlier $B_j$. Thus
$$
\boxed{B_n\cap B_m=\varnothing\ (n\ne m),\qquad B_n\in U_m\Longleftrightarrow m=n.}
$$
Only finite intersections were used; no countable-intersection closure of an <ultrafilter> is being assumed.
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