Prove that each person in a group of five must have exactly two friends under given conditions. 
In a group of five people any two are either friends or enemies, no three of them are friends to each other and no three of them are enemies to each other. Prove that every person in this group has exactly two friends, and hence exactly two enemies.

In this question, the OP is asked to try to solve it using graph theory. Now I am not sure whether I am using proper graph theory or not as I haven't studied it but have only heard of it but I think that my method slightly uses graph theory. So I want to verify my method.
My Approach:
Let us denote the system using a graph. ($A,B,C,D,E$ denote the group of five people)
Let us denote friendship by simple line and enemyship by dotted line.
Now since no three of them are either friends or enemies to each other, a triangle with all its edges of same type is prohibited.

Contrary to what we require to prove, let us assume that $A$ has $4$ friends.

Now $B$ cannot be friends to $D$ or $E$ without forming the prohibited simple triangle, hence he is enemies to both. Now it is evident that $D$ and $E$ can neither be friends nor be enemies, which is a contradiction.
Suppose $A$ has $3$ friends. Two cases are possible for this scenario.
Case I:

WLOG, let $A$ not be friends with $C$.
Now $B$ cannot be friends to $D$ or $E$ without forming the prohibited simple triangle, hence he is enemies to both. Now it is evident that $D$ and $E$ can neither be friends nor be enemies, which is a contradiction.
Case II:

WLOG, let $A$ not be friends with $E$.
Now $B$ cannot be friends to $C$ or $D$ without forming the prohibited simple triangle, hence he is enemies to both. Now it is evident that $C$ and $D$ can neither be friends nor be enemies, which is a contradiction.
Thus $A$ cannot have more than $2$ friends. Similar argument can be used to prove that $A$ cannot have more than $2$ enemies. This means that $A$ must have exactly $2$ friends.
Since $A$ is any arbitrary member of the group, each member of the group must have exactly two friends.
Please check my solution and offer suggestions. Also please provide alternate solutions (limited to high school mathematics) if possible.
THANKS
 A: The general idea of this is exactly correct.
There are a few flaws in it, for instance where you break into cases for $A$ having exactly $3$ friends you only cover the case where $A$ is not friends with $C$ and the case where $A$ is not friends with $E$, but have missed the cases where it was $B$ or $D$ that $A$ was not friends with.
You can get around having to break this into cases however by using the magical phrase "Without Loss of Generality" (abbreviated as "WLOG").  This phrase can be used to signify that whatever possible assumptions follow, the proofs for the other possibilities are analogous to the one provided with the exception of names of variables being replaced and so need not be repeated.
Further, when talking about the case where $A$ had exactly four friends, you never even made use of $C$ so you could have gotten rid of this section of the proof entirely and just rephrased the following section to "$A$ has at least three friends."
So, you could have said, "Suppose that $A$ has at least three friends.  WLOG suppose they were $B,C,D$.  But then to avoid a triangle of friends we know that none of these could have been friends with one another, but then this forms a triangle of enemies... The case of $A$ having at least three enemies follows similarly."
