So the question goes as follows: Let $G_1=(V, E_1)$ and $G_2=(V, E_2)$ be spanning subgraphs of a complete bipartite graph $K_{n,n}$ such that $E_1 \cap E_2 = \emptyset$ and $E_1 \cup E_2 $ contains all edges of $K_{n,n}$
For $n \ge 7$ show that, at least one of $G_1$ and $G_2$ is not planar.
So, I know that $K_{n,n}$ is not planar, and I tried to assume the opposite but it got me nowhere really.
Any help would be appreciated.
$$\max\{\,|E_1|, |E_2|\,\} \ge \frac{|E|}{2} = \frac{n^2}{2}.$$ Suppose that $|E_1| \ge |E_2|$, so $|E_1| \ge \frac{n^2}{2}$.
For any bipartite planar graph $H$ on at least 3 vertices it is true that $m' \le 2n' - 4$, where $n'$ is the number of vertices in $H$ and $m'$ is the number of edges in $H$. This easily follows from Euler's formula $n' - m' + f' = 2,$ where $f'$ is the number of faces of $H$, and the fact that each cycle (and therefore each face) in bipartite planar graph has at least 4 edges while each edge belongs to at most 2 faces: $2m' \ge 4f'$.
So if $G_1$ is planar we have $\frac{n^2}{2} \le |E_1| \le 2\cdot 2n - 4$, then $n^2 - 8n + 8 \le 0$. We see that $n^2 - 8n + 8 > n^2 - 8n + 16 - 9 = (n - 4)^2 - 9 \ge 0$ for $n \ge 7$, so $G_1$ can't be planar.
Hint: what is the maximum number of edges a planar bipartite graph can have?
Euler's formula can be used to show a planar bipartite graph has at most $2n-4$ edges. Thus $G_1$ and $G_2$ have at most $4n-8$ edges in total if they are planar, but this is not enough if $n\geq 7$.