Each set of a finite family of subsets of a line is a union of two closed intervals. Moreover, any three of the sets of the family have a point in common. Prove that there is a point which is common to at least half the sets of the family.
Let these subsets be $S_1, S_2,\ldots,S_n$. Then we can write $S_i=[a_i, b_i]\cup [c_i, d_i]$ where $a_i\le b_i\le c_i\le d_i$. Now, let $a=\max(a_i)$ and $d=\min(d_i)$. Then suppose $a\in S_h$ and $d\in S_k$, and suppose there exists some $S_i$ such that $a\notin S_i$ and $d\notin S_i$. Since $b_i<a$, if there is a point $P$ such that $P\in S_i$ and $P\in S_h$, then $P\notin [a_i, b_i]$. Similarly, since $c_i >b$, if there is a point $Q$ such that $Q\in S_i$ and $Q\in S_k$, then $Q\notin [c_i, d_i]$. But this means $S_i\cap S_h\cap S_k=\emptyset$, which is a contradiction. It follows that every $S_i$ contains one of $a$ or $d$, so either $a$ or $d$ belongs to at least half of these subsets.