Here is Prob. 12, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable.
My Attempt:
Let $D$ be a countable dense subset in the topological space $X$. Then $D \subset X$ and $\overline{D} = X$.
Let $\mathscr{A}$ be a (pairwise) disjoint collection of open sets of $X$.
For each $A \in \mathscr{A}$, we must have $$ A \cap D \neq \emptyset, $$ by Theorem 17.5 (a) in Munkres; let us choose a point $x_A \in A \cap D$.
Now if $A_1$ and $A_2$ are any two distinct sets in $\mathscr{A}$, then since $A_1 \cap A_2 = \emptyset$, therefore we must also have $$ \left( A_1 \cap D \right) \cap \left( A_2 \cap D \right) = \left( A_1 \cap A_2 \right) \cap D = \emptyset, $$ which implies that $$ x_{A_1} \neq x_{A_2}. $$ [Please refer to the preceding paragraph.]
Thus the map $\mathscr{A} \rightarrow D$, $A \mapsto x_A$ is an injection of $\mathscr{A}$ into the countable set $D$. Therefore the collection $\mathscr{A}$ must be countable.
Is this proof correct and satisfactory enough? Or, are there any lacunas?
Here I have assumed that our collection of open sets is pairwise disjoint.
What if our collection is disjoint in the sense that $$ \bigcap_{A \in \mathscr{A} } A = \emptyset, $$ but there exist sets $A, A^\prime \in \mathscr{A}$ such that $A \cap A^\prime \neq \emptyset$? Does the conclusion still hold?
In answer to the the question in the last sentence, let $$ \mathscr{A}=\{B(0,r)|r\in\mathbb{R}\} \cup \{B(1,1/2)\}$$ where $B(x,r)$ is the open disk of radius $r$ centered at $x$ in the plane.
Then the intersection of $\mathscr{A}$ is empty and $\mathscr{A}$ in uncountable.