By merging two-two color classes, the Four Color Theorem implies that every planar graph can be two-colored such that each color class induces a triangle-free graph.
Is there a simpler proof for this fact?



Is there easy proof for triangle-free two-coloring of planar graphs?
Thomassen does indeed prove the vertex version, but [in a different paper](http://www.sciencedirect.com/science/article/pii/S0095895608000336). In fact this paper proves the stronger statement that you can get a coloring without monochromatic triangles from any 2-list coloring.


A bit of googling revealed that originally the claim of the OP was proved in 



> 
> Burstein M.I., "The bicolorability of planar hypergraphs", 
>  Sakharth. SSR Mecn. Akad. Moambe 78 (1975), no. 2, 293–296. (MR0396314)
> 
> 
> 


This paper proves that for any planar triangulation, and any assignment of colors to the outer triangle which is not monochromatic one can extend the coloring to the entire triangulation while avoiding monochromatic triangles.


