Proof by Contradiction--Product of Lindelöf Spaces is Lindelöf 
Consider the three statements:

*

*(i) If $X$ is a set and $\mathcal{A}$ is a collection of subsets of $X$ having the countable intersection property, then there is collection $\mathcal{D}$ of subsets of $X$ such that $\mathcal{D} \supset \mathcal{A}$ is maximal with respect to the countable intersection property.


*(ii) Suppose that $\mathcal{D}$ is maximal with respect to the countable intersection property. Then countable intersections of elements in $\mathcal{D}$ are in $\mathcal{D}$. Furthermore, if $A \subseteq X$ intersects every element of $\mathcal{D}$, then $A \in \mathcal{D}$.
Show that products of Lindelöf  spaces are Lindelöf, and then show that (ii) holds.

Okay. This problem is giving some trouble. It's clear that the intention of this problem is to show (i) is false by reductio, and this can be adequately done by considering just the product of two Lindelöf spaces (since I have a counterexample for this case). In the problem before this, I proved that $X$ is a Lindelöf space if and only if for every $\mathcal{A} \subseteq \mathcal{P}(X)$ having the countable intersection property (CIP), it follows that $\bigcap_{A \in \mathcal{A}} \overline{A} \neq \emptyset$, so I'm going to try using this in proving $X \times Y$ is Lindelöf, where each factor is Lindelöf.
Suppose that $\mathcal{A} \subseteq \mathcal{P}(X \times Y)$ has the CIP, but $\bigcap_{A \in \mathcal{A}} \overline{A} = \emptyset$. Then there exists $\mathcal{D} \supset \mathcal{A}$ which is maximal with respect to the CIP. If all my calculations are right, both $\{\pi_1 (A) \mid A \in \mathcal{A} \}$ and $\{\pi_2(A) \mid A \in \mathcal{A} \}$ have the CIP, which means there exists $\mathcal{D}_1 \subseteq \mathcal{P}(X)$ and $\mathcal{D}_2 \subseteq \mathcal{P}(Y)$ containing the respect set which are maximal with respect to the CIP....
At this point, I don't know what to do. My whiteboard is messy with many failed attempts...I was thinking that considering the set $\{D_1 \times D_2 \mid D_i \in \mathcal{D}_i \}$ might lead to some desired contradiction, perhaps contradicting the fact that $\mathcal{D}$ is maximal in $X \times Y$, but I presently cannot see how. I could use some help.
 A: Products of Lindelöf spaces are not Lindelöf, as the classical example of $X= \mathbb{R}$ with the Sorgenfrey topology (with base of all sets of the form $[a,b)$, $a < b$), and where $X \times X$ is not Lindelöf (as $F = \{(x,-x): x \in \mathbb{R}\}$ is closed in $X \times X$ and discrete in the subspace topology.
(i) will not hold: take the set of co-countable subsets on an uncountable set $X$. This has the CIP. If we could extend it to a maximal such family, we'd have a non-principal countably complete ultrafilter, which implies (Ulam's theorem IIRC) that $|X|$ is a measurable cardinal. There are plenty of uncountable non-measurable cardinals (like $\aleph_1$) so (i) is definitely false.
(ii) might well hold, if we had such a family maximal w.r.t. CIP it probably will be closed under countable intersections (we could add them, contradicting maximality, otherwise). Also if $A$ intersects all members we can add it and get a larger such collection, again contradicting maximality. You should check the details for (ii).
