, 1997

We consider two cases.

Case 1. There exist two $3$-cliques sharing two common people.
Suppose the two $3$-cliques are $\{A, B, C\}$ and $\{A, B, D\}$. If all $3$-cliques contain $A$ or $B$, we are done as we can remove $A$ and $B$. If there exists a $3$-clique without $A$ and $B$, it must be $\{C, D, E\}$ since every pair of $3$-cliques has a common person. We claim that there is no $3$-clique if $C$ and $D$ are removed.
Indeed, note that $\{A, C, D\}$, $\{B, C, D\}$ and $\{C, D, E\}$ are existing $3$-cliques since all pairs of people in these $3$-cliques know each other from above. Therefore, the only possible $3$-clique without $C$ and $D$ is $\{A, B, E\}$. Then $\{A, B, C, D, E\}$ forms a $5$-clique, which is a contradiction.

Case 2. Any pair of $3$-cliques shares exactly one common person.
Suppose two $3$-cliques are $\{A, B, C\}$ and $\{A, D, E\}$. If all $3$-cliques contain $A$, we are done as we can remove $A$. If there exists a $3$-clique without $A$, it must be $\{B, D, F\}$ up to renaming of the people. Then we have a $3$-clique $\{A, B, D\}$ sharing two common people with $\{A, B, C\}$. This is a contradiction.
The proof is complete since all cases are covered.