A society has to elect a board of governors. A society has to elect a board of governors. Each member of the society has chosen $10$ candidates for the board, but he will be happy if at least one of them will be on the board. For each six members of the society there exists a board consisting of two persons making all of these six members happy. Prove that a board consisting of $10$ persons can be elected making every member of the society happy.
Comment: I could easily find a board of $20$ governors. Suppose that the claim does not hold. Take any set $ \mathcal{G} =\{G_1,G_2,...G_{10}\}$ which makes happy member $M_1$. Since the claim does not hold there must be a member $M_2$ who is unhappy with that board, but there is board $\mathcal{G'} =\{G'_1,G'_2,...G'_{10}\}$ which does makes him happy. It is not difficult to see, because of the second assumption, that the board $ \mathcal{G} \cup \mathcal{G'}$ makes the whole society happy.
That is it. I can't go any further.    
 A: Let $A = [a_1,a_2...a_{10}]$ be the set of people that make $P_A$ happy. Then assume there exists $P_B$ (with set $B = [b_1,b_2...b_{10}])$ who is not happy with $A$ as the board and then as you've shown $A\cup B$ makes everyone happy.
Then assume $A\cup B$ is the minimum sized board st everyone is happy. Therefore there exists $C$ st $C \cap (A\cup B) = a_1$ as otherwise $a_1$ could be removed and everyone would still be happy with the board. By similar logic there exists $D$ st $D \cap (A\cup B) = a_2$. However if we put $A,B,C$ and $D$ in a group of $6$ there isn't a board of $2$ that makes all of them happy. Therefore we can remove either $a_1$ or $a_2$ and everyone would still be happy. Say we have just removed $a_1$ then there exists $E$ st $E \cap (A\cup B) = a_3$ or $[a_1,a_3]$ and there exists $F$ st $F \cap (A\cup B) = a_4$ or $[a_1,a_4]$ so therefore we can either remove $a_3$ or $a_4$ (using the same argument as before) or remove $a_3$ and $a_4$ and put $a_1$ back in, reducing the size of the board whatever the case. 
We can continue this process removing all but one of $[a_1,a_2...a_{10}]$ and all but one of $[b_1,b_2...b_{10}]$ leaving a board of $2$. (I think there are still some problems with this argument). Therefore if there are $2$ people whose choices don't have anyone in common a board of $2$ is enough to make everyone happy and if not then $[a_1,a_2...a_{10}]$ makes everyone happy so a board of $10$ people is enough whatever.
