TITLE: A problem in topology relating to the finite intersection property
QUESTION [5 upvotes]: This is a problem from Munkres' Topology Exercise 37.1 (c)
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.

(c) Show that if $X$ satistifes the T1 axiom, there is at most one point belonging to $\bigcap_{D\in \mathscr{D}}\bar D$.

The property that I'm attempting to use is the following Lemma.
Lemma 37.2

Let $X$ be a set; let $\mathscr{D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property. Then:
If $A$ is a subset of $X$ that intersects every element of $\mathscr{D}$ then $A$ is an element of $\mathscr{D}$.

To use this property, I've been trying to show that {$\bar D | D \in \mathscr{D}$}is equal to $\mathscr{D}$, however, I'm stuck here as you can see below. How can I work this out? Here is what I have:

To lead to a contradiction, assume there are distinct $x_1,x_2\in\bigcap_{D\in\mathscr{D}}\overline D$. Then we have $x_1,x_2\in\overline D$ for all $D\in\mathscr{D}$ and so $\{x_1\},\{x_2\}\subset\overline D$ for all $D\in\mathscr{D}$. By $T_1$, $\{x_1\}$ and $\{x_2\}$ are closed in $X$.
Let $\mathscr{D}'=\{D\mid D\in\mathscr{D}\}\cup\{\overline D\mid D\in\mathscr{D}\}$. Then $\mathscr{D}\subset\mathscr{D}'$. Claim: $\mathscr{D}'$ has f.i.p.
Take any finite elements $C_1,\ldots,C_n\in\mathscr{D}'$.
i. if all $C_1,\ldots,C_n\in\mathscr{D}$, then $C_1\cap\ldots\cap C_n\ne\varnothing$ by f.i.p. of $\mathscr{D}$.
ii. if all $C_1,\ldots,C_n\in\{\overline D\mid D\in\mathscr{D}\}$, then $C_1\cap\ldots\cap C_n\ne\varnothing$ since $x_1,x_2\in\bigcap_{D\in\mathscr{D}}\overline D$.
iii. if $D_1,\ldots,D_k\in\mathscr{D}$, $\overline{D_{k+1}},\ldots,\overline{D_n}\in\{\overline D\mid D\in\mathscr{D}\}$, $$D_1\cap\ldots\cap D_k\cap\overline{D_{k+1}}\cap\ldots\cap\overline{D_n}\supset D_1\cap\ldots\cap D_n\ne\varnothing$$ by f.i.p. of $\mathscr{D}$. Thus by maximality, $\mathscr{D}=\mathscr{D}'$, so $\{D\mid D\in\mathscr{D}\}\supset\{\overline D\mid D\in\mathscr{D}\}$.

REPLY [2 votes]: Here I assume $X$ is T$_2$.
Suppose $x,y\in X$ and $x\neq y$. Let $U,V\subseteq X$ be disjoint open sets containing $x$ and $y$, respectively. Then either $U$ or $X\setminus U$ is in $\mathcal D$ (can you prove this?).  If $X\setminus U \in \mathcal D$ then $x\notin\bigcap_{D\in\mathcal D} \overline D$.  If $U\in\mathcal D$ then since $\overline U$ misses $y$ we have $y\notin \bigcap_{D\in\mathcal D} \overline D$.