A trivial but efficient numerical proof without assuming that friends of my friends are my friends:
Each pair know a third fellow. There are \(\binom 5 2 = 10\) pairs \(\equiv 20\) arrows starting from each to another element ( no self arrow allowed ).
The mean of received arrows is \(2 \frac{\binom 5 2}5 = 2 \frac{10}5 = 4\).
Receiving 4 arrows means the receiver knows the other \(4 \equiv\) everybody.
It is a mean. If you decrease the arrows received by someone , you must increase the arrows received by another one.
Hence, there is always someone receiving at least 4 arrows. Then there is always someone which is the friend of "everybody".
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