A set of sets A is said to be disjointed if $\forall C,D\in A, C\cap D =\varnothing $. Let F be a set of sets; prove that F has a maximal disjointed subset. Prove that this statement is equivalent with the Axiom of Choice.
IDEA
My idea is to use Zorn's Lemma that says: Every inductive set has at least one maximal element.
For the direction not covered by Zorn's Lemma, suppose we're given that every family of sets has a maximal disjointed subfamily, and suppose we're given a family $\mathcal F$ of nonempty sets; we intend to find a choice function for the family.
To simplify notation, assume, without loss of generality, that $\mathcal F$ is disjointed. (If it's not, replace each $A\in\mathcal F$ with $\{\langle A,a\rangle:a\in A\}$.)
For each given element $a\in\bigcup\mathcal F$, let $A_a$ be the element of $\mathcal F$ that contains $a$ (it's unique because $\mathcal F$ is disjointed), and let $Z_a$ be the set of all the sets $\{a,x\}$ where $a$ is our given element and $x\in A_a$. (So in nontrivial situations most of the elements of $Z_a$ will be $2$-element sets, but we also allow $x=a$.) Notice that $Z_a$ and $Z_b$ will be disjoint iff $A_a\neq A_b$, because if $A_a=A_b$ then $\{a,b\}$ is in both $Z_a$ and $Z_b$.
By hypothesis, $\{Z_a:a\in\bigcup\mathcal F\}$ has a maximal disjointed subfamily $\mathcal D$. By the observation at the end of the preceding paragraph, the set $C=\{a\in\bigcup\mathcal F:Z_a\in\mathcal D\}$ contains at most one $a$ from each $A\in\mathcal F$. Maximality easily implies that $C$ actually contains at least one $a$ from each $A\in\mathcal F$, because if it missed some $A\in\mathcal F$ then we could take an arbitrary $a\in A$ (remember that the sets in $\mathcal F$ are nonempty) and add $Z_a$ in to $\mathcal D$, contradicting maximality.
So $C$ contains exactly one element from each $A\in\mathcal F$ and we get the desired choice function by sending each $A\in\mathcal F$ to its unique element in $C$