On pairwise non-disjointness of a sequence of finite sets whose sizes are uniformly bounded Let $\{A_n\}_{n=1}^\infty $ be a sequence of finite sets such that $\sup \{|A_n|: n \in \mathbb N\}$ is finite and no two sets of the sequence are pairwise disjoint ; then must there exist a finite set $F$ such that no two sets in the sequence $\{ A_n \cap F\}_{n=1}^\infty $ are pairwise disjoint ?
related On pairwise non-disjointness of a sequence of finite sets
 A: I had to prove a more general statement to make the induction work. If $\mathcal A,\mathcal B$ are families of sets, let me say that "$\mathcal A$ meets $\mathcal B$" if $A\cap B\ne\emptyset$ whenever $A\in\mathcal A,\ B\in\mathcal B,$ and that "$\mathcal A$ meets $\mathcal B$ in $F$" if $A\cap B\cap F\ne\emptyset$ whenever $A\in\mathcal A,\ B\in\mathcal B.$ The following theorem reduces to the statement you asked about when $\mathcal A=\mathcal B=\{A_1,A_2,A_3,\dots\}.$
Theorem. Let $n$ be a positive integer. Suppose $\mathcal A,\mathcal B$ are families of finite sets such that $\mathcal A$ meets $\mathcal B.$ If every member of $\mathcal A\cup\mathcal B$ has size $\le n,$ then there is a finite set $F$ such that $\mathcal A$ meets $\mathcal B$ in $F.$
Proof by induction on $n.$ The case $n=1$ is trivial. Suppose $n\gt1.$ The statement is vacuous if $\mathcal A$ or $\mathcal B$ is empty, so we assume they are both nonempty. Choose $A_0\in\mathcal A$ and $B_0\in\mathcal B,$ and let $E=A_0\cup B_0.$ Now every member of $\mathcal A\cup\mathcal B$ has nonempty intersection with $E.$
For each nonempty set $X\subseteq E$ let
$$\mathcal A_X=\{A\in\mathcal A:A\cap E=X\},\ \mathcal B_X=\{B\in\mathcal B:B\cap E=X\}.$$
Then $\mathcal A=\bigcup_{\emptyset\ne X\subseteq E}\mathcal A_X,\ \mathcal B=\bigcup_{\emptyset\ne Y\subseteq E}\mathcal B_Y.$ It will suffice to find, for each pair $X,Y$ of nonempty subsets of $E,$ a finite set $F_{X,Y}$ such that $\mathcal A_X$ meets $\mathcal B_Y$ in $F_{X,Y};$ then we can take the union of those sets $F_{X,Y}$ to be our $F.$ If $X\cap Y\ne\emptyset$ we can simply take $F_{X,Y}=E.$
If $X\cap Y=\emptyset,$ then $\mathcal A'_X=\{A\setminus E:A\in\mathcal A_X\}$ meets $\mathcal B'_Y=\{B\setminus E:B\in\mathcal B_Y\};$ since every member of $\mathcal A'_X\cup\mathcal B'_Y$ has size $\le n-1,$ we can use the inductive hypothesis to get a finite set $F_{X,Y}$ such that $\mathcal A'_X$ meets $\mathcal B'_Y$ (and so $\mathcal A_X$ meets $\mathcal B_Y$) in $F_{X,Y}.$
