A Question regarding disjoint dense sets If we take the standard topology on $\mathbb{R}$ we can easily find two disjoint sets that are dense, namely $\mathbb{R}\setminus\mathbb{Q}$ and $\mathbb{Q}$. Similarily, if we take the same topology and restrict it to $\mathbb{Q}$ we can again find two disjoint dense sets.
Define for every prime $p$: $$A_p:=\left\{ \frac{i}{p^j} : i,j\in\mathbb{Z}\right\} \setminus \mathbb{Z}$$
For any pair of distinct primes $p, q$ one can easily see that $A_p \cap A_q=\emptyset$, and both $A_p$ and $A_q$ are dense.
This made me curious about general topologies, and I attempted to prove or disprove the following hypothesis:
Given any topological space $(X,\mathcal{T})$, such that for all non-empty $S\in \mathcal{T}$ we have $|S| \ge2$, there exist two disjoint dense subsets of $X$.
My Initial Attempt:
Use the axiom of choice to define a function $f$ that selects an item from each non-empty open set in $X$. Now, use the axiom again of define a function $g'$ that selects an item from each set in:
$$
\mathcal{T}':=\left\{ S\setminus\left\{f(S)\right\}\,:\,S\in\mathcal{T}\setminus\{\emptyset\} \right\}
$$
and then define $g(S):=g'\left(S\setminus\left\{f(S)\right\}\right)$ for all non-empty $S\in \mathcal{T}$.
This means that both $f(\mathcal{T})$ and $g(\mathcal{T})$ are dense, and moreover $f(S) \neq g(S)$ for all non-empty $S\in \mathcal{T}$. My attempt was to show that $f(\mathcal{T})$ and $g(\mathcal{T})$ are disjoint, but I later realized that this is incorrect. Any help will be very appriciated.
 A: In technical terms, your question can be stated as "Is every perfect space resolvable?" 
The answer to that is no.
A standard counterexample is a space $(X, T)$ where $T \setminus \{ \emptyset \}$
is a free ultrafilter on $X$.
Since the ultrafilter is free, all nonempty open sets are infinite.
Because it is an ultrafilter every nonempty proper subset is either open or closed, and therefore any dense proper subset must be open. On the other hand,
no two sets in a filter are disjoint, and it follows that no two dense subsets
can be disjoint. Note that the last part works more generally for any "door space".
A: I'm going to assume the space has a countable basis consisting of open sets with uncountable cardinality.
Choose a point from every basis set to get a dense set $A$. Now if there is no open set such that all of its points have been chosen, we are done, just choose the complement of $A$. But because we have just countably many points, not every point in each chosen set has been chosen.
Edit: Ok, this case was simpler than I thought. And not even very interesting.
