Bonus property of dense set decomposition? Suppose a topological space X can be written as the union of two dense sets A and B, i.e. both intersect every open $\emptyset\neq U\subseteq X$. Supposing X has no finite open sets (but $\emptyset$), can A and B be always chosen such that they intersect each U on a set the size of U?
 A: A partial result:
Suppose $X = A \cup B$ where $A,B$ are infinite, dense and disjoint.
I claim that $A$ and $B$ can also be chosen to have the same cardinality.
Suppose $\kappa= |A| > |B|$, then partition $A$ into $\kappa$ many disjoint sets of size $\kappa$, say $A = \cup_{\alpha < \kappa} A_\alpha$. 
Suppose that none of the sets $A \setminus A_\alpha$ is dense in $X$. Then for every $\alpha <\kappa$ there is some $b_\alpha \in B$ such that $b_\alpha \notin \overline{A \setminus A_\alpha}$ (or else $B \subseteq \overline{A \setminus A_\alpha}$, which would make $A \setminus A_\alpha$ dense...). The map from $\kappa$ to $B$ defined by $\alpha \rightarrow b_\alpha$ is not injective (as $|B| < \kappa$), so for some $\alpha, \beta < \kappa$ we have $b = b_\alpha = b_\beta$. But then $\overline{A} = \overline{A \setminus (A_\alpha \cap A_\beta)} = \overline{A\setminus A_\alpha} \cup \overline{A \setminus A_\beta}$ does not contain $b$, contradicting the denseness of $A$.
So some $A \setminus A_\alpha$ is dense. But then we can write $X$ as the disjoint union of $A \setminus A_\alpha$ (of size $\kappa$) and $B \cup A_\alpha$ of size $\kappa$. 
So for the open set $X$ the answer is yes...
This idea might be extensible to other open sets as well. 
