discrete mathematicsgraph theory
Consider of graph $G$ being the union of two simple planar graph both on the same set of vertices. I want to show $\chi(G) \leq 12$.
Four colour theorem indicates that the chromatic number of a planar graph is less than 4, and for general graphs $\chi(G1 \cup G2) \leq \chi(G1)*\chi(G2)$. But this would only yield an upper bound of 16. Is there anything particular to planar graph that helps to reduce the bound?
The inequality $|E(G)| \le 3 |V(G)| - 6$ is a more useful direction to go in here than the $4$-color theorem. The proof of this result is more similar to the proof of the $6$-color theorem which, in summary, goes as follows:
This argument generalizes to an argument based on the degeneracy of a graph. A $k$-degenerate subgraph is one in which every subgraph has a vertex of degree at most $k$. The inequality above shows that any planar graph is $5$-degenerate, and a similar inductive argument shows that all $k$-degenerate graphs are $(k+1)$-colorable.
So if you prove that the union of $100$ planar graphs is $599$-degenerate, you'll have a proof that it's $600$-colorable.
In general it does not. Let
By your definition of union, $$V(G_2\cup G_3)=\{1,2,3,4,5\}\quad\text{ and }\quad E(G_2\cup G_3)=\{(1,2)\}$$ So that $$\chi(G_2\cup G_3)=\chi(G_2)=2$$
However $G_3\not\subseteq G_2$.
You need additional restriction.
Edit - Non trivial example Let
Note that $G_3\not\subseteq G_5$ but $\chi(G_5\cup G_3)=\chi(G_5)=3$$
Edit 2 Anticipating additional question, the results does not hold even if $V(G_p)\not\subseteq V(G_q)$ and $V(G_q)\not\subseteq V(G_p)$ and $V(G_p)\cap V(G_q)\neq\emptyset$ :
Again that $G_3\not\subseteq G_5$ but $\chi(G_5\cup G_3)=\chi(G_5)=3$$
Best Answer
Planar graphs satisfy $|E(G)| \le 3|V(G)|-6$, so the union of two planar graphs satisfies $|E(G)| \le 6|V(G)|-12$, which we can use to show that there is a vertex of degree less than $12$.
This leads us to a proof by induction: remove that vertex, color the rest, and then color that vertex. The inductive hypothesis applies because any subgraph of $G$ is also a union of two planar graphs: the corresponding subgraphs of $G_1$ and $G_2$.