Solution:

If $P$, $Q$, $R$, $S$, and $T$ are any five distinct players, then consider all pairs $A$, $B \in \{P, Q, R, S, T\}$ such that $A$ takes lessons from $B$. Each pair contributes to exactly three triples $(A, B, C)$ (one for each of the choices of $C$ distinct from $A$ and $B$); three triples $(C, A, B)$; and three triples $(B, C, A)$. On the other hand, there are $5 \times 4 \times 3 = 60$ ordered triples of distinct players among these five, and each includes exactly one of our lesson-taking pairs. That means that there are $60 / 9$ such pairs. But this number isn't an integer, so there cannot be five distinct people in the club.

On the other hand, there can be four people, $P$, $Q$, $R$, and $S$: let $P$ and $Q$ both take lessons from each other, and let $R$ and $S$ both take lessons from each other; it is easy to check that this meets the conditions. Thus the maximum number of players is $4$.