Suppose the members are $A, B, C, D, E, F$. By the pigeonhole principle, either $A$ knows at least $\left\lfloor \frac{5}{2} \right\rfloor = 3$ other members, or $A$ does not know at least 3 other members. WLOG assume it is the former case. Suppose $A$ knows $B, C, D$. If any two of $B, C, D$ know each other, say $B$ and $C$, then we are done since all of $A, B, C$ know each other. Otherwise, any two of $B, C, D$ do not know each other, which is also our goal. This completes the proof.