Lindelöf if and only if every collection with the countable intersection property has non-empty intersection of closures I am trying to study for my topology exam, and my professor recommended this question from the text (Munkres's Topology (2nd edition), Section 37 question 2):
A collection $\mathcal{A}$ of subsets of $X$ has the countable intersection property if every countable intersection of elements of $\mathcal{A}$ is nonempty. Show that $X$ is a Lindelöf space if and only if, for every collection $\mathcal{A}$ of subsets of $X$ having the countable intersection property, $\bigcap\limits_{A \in \mathcal{A}} \overline{A} \ne \emptyset$.
 A: I think it is easier to assume that the proposition is false in both directions to get a contradiction.
First, lets suppose that $X$ is Lindelöf but there's a collection $\mathcal{A}$ of subsets of $X$ with the countable intersection property such that $\bigcap_{A\in \mathcal{A}}\bar{A}=\emptyset$. Then, the collection $\mathcal{O}=\{X\setminus\bar{A}:A\in\mathcal{A}\}$ is an open cover of $X$. If follows that there's a countable subcover $\mathcal{O}'=\{X\setminus\bar{A_1}, X\setminus\bar{A_2},\ldots\}$, that is , $X=\bigcup_{n=1}^\infty X\setminus\bar{A_n}$.
But this implies that
$$
\emptyset=X\setminus \bigcup_{n=1}^\infty X\setminus\bar{A_n}=\bigcap_{n=1}^\infty \bar{A_n}
$$
Since for every positive integer, $A_n\subset \bar{A_n}$, we have that $\bigcap_{n=1}^\infty A_n=\emptyset$. This contradicts that $\mathcal{A}$ has the countable intersection property.
Next, assume that $X$ has the stated property but it is not Lindelöf. This means that there's an open cover $\mathcal{O}$ such that there's no countable subcover. Consider the collection of closed sets $\mathcal{A}=\{X\setminus U:U\in\mathcal{O}\}$. Note that this collection has the countable intersection property, for if a countable intersection of elements in $\mathcal{A}$ is empty, $\bigcap_{n=1}^\infty X\setminus U_n=\emptyset$, then
$X=X\setminus \bigcap_{n=1}^\infty U_n=\bigcup_{n=1}^\infty U_n$ and $\mathcal{O}$ would have a countable subcover. If follows, by assumption, that $\bigcap_{A\in\mathcal{A}}\bar{A}=\bigcap_{A\in\mathcal{A}}A \neq \emptyset$ (the first equality is because every set in $\mathcal{A}$ is closed).Since
$$X\setminus \bigcap_{A\in\mathcal{A}}A=X\setminus\bigcap_{U\in \mathcal{O}}(X\setminus U)=\bigcup_{U\in\mathcal{O}}U$$
This means that $\mathcal{O}$ is not a cover, a contradiction.
