Of all possible groups of people we can form in this class, we consider the groups in which everyone knows each other, and from these we take one with as many people as possible. (This largest group exists: there are finitely many people and there is at least a group of people who all know each other, namely a group consisting of one person.) Call this group $X$. We prove that the group $X$ and the group consisting of the rest of the people satisfy the condition.
Since everyone in $X$ knows each other, we only need to prove that all people outside $X$ do not know each other. Suppose there are $A$ and $B$ outside $X$ who do know each other. Since $X$ is as large as possible, $X$ contains a person $A^{\prime}$ who is not an acquaintance of $A$. Similarly, $X$ contains a person $B^{\prime}$ who is not an acquaintance of $B$. We now first prove that we can choose $A^{\prime}$ and $B^{\prime}$ to be different from each other. If not, there is a unique person $C=A^{\prime}=B^{\prime}$ in $X$ who does not know $A$ and $B$. For all other people in $X$, it then holds that they know both $A$ and $B$. Now consider the set $(X \backslash C) \cup\{A, B\}$. In this group of people, everyone knows each other: the people from $X$ know each other, $A$ and $B$ know each other, and everyone in $X$ knows both $A$ and $B$. This group is, however, larger than $X$, a contradiction.
Therefore, we may assume that $A^{\prime}$ and $B^{\prime}$ are different people. Note that $A^{\prime}$ and $B^{\prime}$ know each other, as they are both in $X$. If now $A, B, B^{\prime}$, and $A^{\prime}$ sit down at a round table in that order, everyone knows exactly one of their neighbors: $A$ and $B$ know each other and $A^{\prime}$ and $B^{\prime}$ know each other, but $A$ and $A^{\prime}$ do not know each other, and the same goes for $B$ and $B^{\prime}$. This is a contradiction to the given condition. We conclude that the division into $X$ and the rest of the people satisfies the condition.