We prove that any simple graph with $2n$ vertices and without triangles has at most $n^2$ edges. We prove this by induction. When $n=1$, clearly there is at most $n^2=1$ edge between 2 vertices. Assume there are at most $k^2$ edges when there are $2k$ vertices. Consider a simple graph with $2(k+1)$ vertices and without triangles.

We are done if there is no edge. So we may assume there is an edge $A-B$. For any other vertex $C$, at most one of the pair $\{A, C\}$ and $\{B, C\}$ is joined by an edge since there is no triangle. Thus, there are at most $2k$ edges containing exactly one of $A$ and $B$. Among the remaining $2k$ vertices, there are at most $k^2$ edges by the inductive hypothesis. Therefore, the number of edges is at most
$$1 + 2k + k^2 = (k+1)^2.$$
This proves the inductive step, and so we are done.