Consider a 6-good club and denote some seven of its members by $A, \dots, G$. It suffices to show that $A, \dots, G$ can be seated around a table as required. Consider only friendships among $A, \dots, G$. First, we show that every member has at least three friends.

Without loss of generality consider $G$. By assumption, $B, \dots, G$ can be seated as required, hence $G$ has at least two friends. Without loss of generality, $F$ is one of them. By assumption, $A, \dots, E, G$ (omitting $F$) can be seated as required, hence $G$ has at least two more friends apart from $F$ for a total of at least three friends.

Since every member has at least three friends, there exists a member with at least four friends (otherwise the number of friendly pairs equals $\frac{1}{2} \cdot 7 \cdot 3$, which is clearly impossible). Without loss of generality, assume $G$ has at least four friends.

By assumption, $A, \dots, F$ can be seated as required. In such a seating, some two of the four friends of $G$ are neighbors and we can seat $G$ in between them.