Solution:

We model the group as a graph where each person is a vertex and an edge connects two people if they are enemies. Each vertex has degree less than $4$.

We need to partition the vertices into two sets so that each vertex has at most one neighbor (enemy) in its own set.

We proceed by induction on the number of people.

Base case: For $1$ or $2$ people, the statement is clear.

Inductive step: Suppose the statement is true for all groups with fewer than $n$ people. Consider a group with $n$ people.

Since every person has less than $4$ enemies, the graph has maximum degree at most $3$.

Pick any person $v$. Remove $v$ from the group. By the induction hypothesis, the remaining group can be partitioned into two sets $A$ and $B$ so that each person has at most one enemy in their own set.

Now, add $v$ back. $v$ has at most $3$ enemies, say $x$, $y$, $z$ (possibly fewer). These enemies are in $A$ and $B$ in some way.

If at most one of $x$, $y$, $z$ is in $A$, put $v$ in $A$; then $v$ has at most one enemy in $A$ (his own set).

If at least two of $x$, $y$, $z$ are in $A$, then at most one is in $B$, so put $v$ in $B$.

Thus, we can always place $v$ in a set so that he has at most one enemy in his own set.

This completes the induction.