Prove existence a maximal pairwise disjoint subfamily $\mathcal{S} \subseteq \mathcal{R}$ Can anyone help me with this:
Suppose $\mathcal{R}$ is any family of sets. Prove that there exists set $\mathcal{S} \subseteq \mathcal{R}$, which is pairwise disjoint and for each $A \in \mathcal{R}-\mathcal{S}$, there exists $B  \in  \mathcal{S}$ with following property: $A  \cap  B \neq \emptyset$. 
I have no idea where to start. Thanks in advance for help.
 A: This looks like a job for Zorn's Lemma! Consider the family $\mathscr F$ of subsets of $\mathcal R$ that are pairwise disjoint. Note that $\mathscr F$ is non-empty and partially-ordered by $\subseteq.$ (Why?)
Now, let $\mathscr C$ be a $\subseteq$-chain of $\mathscr F.$ That is, suppose that $\mathscr C$ is a subfamily of $\mathscr F,$ and that $\subseteq$ is a non-strict total order on $\mathscr C.$ Put $\mathcal C=\bigcup\mathscr C.$ As a union of subsets of $\mathcal R,$ we have that $\mathcal C$ is likewise a subset of $\mathcal R.$ Take any $A,B\in\mathcal C,$ meaning that $A\in\mathcal C_1$ and $B\in\mathcal C_2$ for some $\mathcal C_1,\mathcal C_2\in\mathscr C.$ Since $\mathscr C$ is totally ordered by $\subseteq,$ then $\mathcal C_1\subseteq\mathcal C_2$ or $\mathcal C_2\subseteq\mathcal C_1,$ so in either case, $\mathcal C_1\cup\mathcal C_2\in\mathscr C,$ and so $A,B\in\mathcal C_1\cup\mathcal C_2\in\mathscr F.$ By definition of $\mathscr F,$ either $A=B$ or $A\cap B=\emptyset.$ Hence, the elements of $\mathcal C$ are pairwise disjoint, so $\mathcal C\in\mathscr F.$ Furthermore, we readily see that $\mathcal D\subseteq\mathcal C$ for any $\mathcal D\in\mathscr C.$
Thus, by Zorn's Lemma, $\mathscr F$ has a $\subseteq$-maximal element, say $\mathcal S.$ Can you take it from there (and justify the claims made before)?

As a side note, we need Zorn's Lemma (or something like it) to prove this result, at least as far as Zermelo-Fraenkel set theory is concerned
