Let us prove that any two friends $A$ and $B$ belong to some special team. Let us define a longest sequence $S_1, S_2, \dots, S_m$ of students such that

* for all $1 \le i < j \le m$ we have $S_i \ne S_j$
* for each $1 \le i \le m$ any team containing $S_i$ also contains $A, B, S_1, S_2, \dots, S_{i-1}$.

Note that the sequence is well defined since it contains at least one element. Indeed, if $A$ and $B$ end their friendship then some student $S_1$ does not belong to any team. Therefore any team containing $S_1$ contains both $A$ and $B$ ($S_1$ may coincide with $A$ or $B$).

Let $T$ be any team of $S_m$. Assume that $S_m$ has a friend $S'$ outside of $T$. If $S_m$ and $S'$ end their friendship then there is a student $S''$ which does not belong to any team. Therefore, any team containing $S''$ contains also $S_m$ and consequently contains $A, B, S_1, S_2, \dots, S_{m-1}$. Note that $S''$ does not coincide with $S_1, S_2, \dots, S_{m-1}$ since any team of $S_m$ contains $S_1, S_2, \dots, S_{m-1}$ and any team of $S''$ contains $S'$. Then $S''$ can be added to the sequence $S_1, S_2, \dots, S_m$. This contradicts the maximality of this sequence. Thus, the team $T$ is special (and it is the only team of $S_m$). Done.