Proving certain set exists from axioms Given the following axiom system:

AXIOM SYSTEM
Definitions:
A committee is a container, possibly empty, of persons.
A person is a member of a committee iff the person is contained in the
committee.
Two committees are disjoint iff there is no person contained in both
committees.
Axioms:
$A_1$: There exists at least one person.
$A_2$: There is a finite number of committees.
$A_3$: Every person is a member of at least one committee.
$A_4$: For every two different persons, there is one and only one
committee of which each person is a member.
$A_5$: For every committee containing at least one member, there is
one and only one other committee containing at least one member such
that the two committees are disjoint.

I want to prove that each person is a member of at least three committees. I've previously proven that there exists at least 4 persons. Unfortunately, I'm a bit stuck. I'm not sure where to start. I've tried a few different things, but nothing has lead me anywhere. Any suggestions?
Note that I'm not looking for a proof, and the smaller the suggestion the better (i.e. I want to do as much of this myself as possible).
 A: Denote people and committees with lower- and upper-case letters. I'll let you spot which axioms I'm using where.
Fix $a,\,A$ with $a\in A$, and $B$ nonempty but disjoint from $A$, say with $b\in B$. We'll prove $a$ is in at least three committees. I'll use arbitrary other sets where necessary.
Some $C$ has $\{a,\,b\}\subseteq C$, but $C$ is neither $A$ nor $B$. Similarly fix nonempty $D$ disjoint from $C$, say with $d\in D$, whence $d\notin\{a,\,b\}$. So some set $E$ has $\{a,\,d\}\subseteq E$, whence $a\in A\cap C\cap E$.
Since $d\in E\implies E\ne C$, the only way for $a$ not to be an element of at least three sets is to have not only $E=A$, but also $\forall d\in D(d\in A)$, so $D\subseteq A\setminus\{a\}$. Since only one nonempty set is disjoint from $B$, $D=A$, a contradiction.
Lastly, note we can't increase this number. If the people form a square's vertices, and the committees form its sides and diagonals, all axioms are satisfied with each person belonging to exactly three committees.
