Let $X$ be a Hausdorff space. Let $\mathcal{D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property. That means that $\mathcal{D}$ is a family of subsets of $X$ that has the finite intersection property such that any other family properly containing $\mathcal{D}$ does not have the finite intersection property. I would like to exhibit the following assertions:
(a) $x \in \overline{D} \ \forall D \in \mathcal{D} \iff$ every neighbourhood of $x$ belongs to $\mathcal{D}$. Which implication uses the maximality assumption?
(b) Let $D \in \mathcal{D}$. Show that: $A \supset D \Rightarrow A \in \mathcal{D}$
(c) If $X$ is $T_1$ there is no more than one point in $\displaystyle \bigcap_{D \in \mathcal{D}} \overline{D}$
I think that the reverse implication of $(a)$ is straightforward. I am only worrying about the forward one, which I assume is the one that uses maximality. I have had no luck with the others.
I realise that there is a conflict. In $(c)$ one must assume that $X$ is $T_1$ but that comes for free since we assume $X$ Hausdorff. Our professor told us that one should assume Hausdorff otherwise the statement is wrong, although he did not specify for which part. The book does not provide the Hausdorff condition.