I'm reading "General Topology" of James Munkres book, chapter 5 about Tychonoff theorem and got stuck on this problem:
Let $X$ be a space. Let $\mathscr{D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property.
a) Show that $x \in \overline{D}$ for every $D \in \mathscr{D}$ iff every neighborhood of $x$ belongs to $D$.
b) Let $D \in \mathscr{D}$. Show that if $D \subset A$, then $A \in \mathscr{D}$
c) If $X$ satisfies the $T1$ theorem, there is at most one point belonging to $\bigcap_{D \in \mathscr{D}}\overline{D}$
I can prove part a) and b), but I got stuck in part c). I assume that part c) should use results from part a) or b).
Suppose $x$ and $y$ both belongs to the intersection in part c), then because X is T1, then there must be neighborhood $U$ of $x$ which does not contain $y$, and neighborhood $V$ of $y$ which does not contain $x$. Due to part a), we can state that both $U, V \in \mathscr{D}$. Up to here, and I don't know how to continue. Please help me or give me some hint. Thanks a lot.
I think c) is false as stated. Take $X$ a countable set in the co-finite topology (which is $T_1$), and $\mathcal{D}$ a free ultrafilter on $X$. Then $\mathcal{D}$ is maximal w.r.t. the f.i.p. and the closure of any $D \in \mathcal{D}$ is $X$ (as the sets are all infinite). So $\cap \{\overline{D} : D \in \mathcal{D} \} = X$, and not a singleton.