Let $\mathcal F$ be a non-principal ultrafilter of subsets of the set $X$. Let $x \in X$.
Does there exist a decreasing sequence $F_1 \supset F_2 \supset ...$ in $\mathcal F$ such that $\bigcap_n F_n = \{x\}?$
This is true if $X = \mathbb N$, with $F_n = \{n, n+1, ...\} \cup \{x\}$. (Every non-principal ultrafilter contains the co-finite filter.) I'm not sure how to settle it for an uncountable $X$ though.
An ultrafilter is countably complete if it's closed under countable intersection. Of course a principal ultrafilter is countably complete. The existence of a countably complete nonprincipal ultrafilter (on some sufficiently large set) can't be proved in ZFC (standard set theory, which is consistent), but it's widely believed that it can't be disproved either.
If the ultrafilter $\mathcal F$ is nonprincipal and countably complete, and if $F_n\in\mathcal F$ for each $n\in\mathcal N$, then $\bigcap_n F_n\ne\{x\}$, since $\bigcap_n F_n\in\mathcal F$ while $\{x\}\notin\mathcal F$.
On the other hand, if the ultrafilter $\mathcal F$ is not countably complete, then you can find:
(1) a sequence of elements of $\mathcal F$ whose intersection is not in $\mathcal F$;
(2) a decreasing sequence of elements of $\mathcal F$ whose intersection is not in $\mathcal F$;
(3) a decreasing sequence of elements of $\mathcal F$ whose intersection is empty;
(4) a decreasing sequence of elements of $\mathcal F$ whose intersection is $\{x\}$.