Chromatic number in a union of planar graphs I am trying to solve the following problem:

Let $G_1, G_2, \dots,G_{100}$ be $100$ planar graphs on the same vertex set $V$ , with edge sets $E_1, E_2,...,E_{100}$, respectively, and consider the graph $G = (V, \bigcup_{i=1}^{100}E_i)$ which is the union of the graphs $G_1, G_2,\dots,G_{100}$. Prove that $\chi(G) ≤ 600$.

I tried to play around with the inequality $E(G) \leq 3 |G| - 6$ that holds for any planar graph, and also with the fact that we can colour any planar graph with 4 colours but I didn't manage to advance after that. I also though of induction on the number of vertices so that I could do that for an arbitrary set of vertices but this didn't help me either. 
Can anyone provide a hint/direction so I can make some progress?

I think I have the answer.
the $G_1, ..., G_{100}$ are planar graphs on the same set of vertices, thus |V| is the same for all $G_i$.
We know  $|E(G_i)|≤3|V|−$6  $\forall i $, since they are all planar thus $ |\bigcup_{i=1}^{100}E_i| \leq |\sum_{i=1}^{100} E_i| ≤100 (3|V|-6)= 300|V|-600$
For convenience let $ E = \bigcup_{i=1}^{100}E_i$
By the degree sum formula or handshake lemma,$\sum_{v} deg(v) = 2|E| \leq 2*(300|V|-600) = 600 |V|-1200 $, thus there exists (proof by contradiction or average degree) a vertex $v$ with degree strictly less than 600, i.e. degree less than or equal to 599. Thus since every subgraph of the union of 100 planar graphs has a vertex of degree less than or equal to 599, the graph is by definition 599- degenerate and therefore 600-colourable.
Is this correct?
