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?
Thomassen does indeed prove the vertex version, but in a different paper. 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.
Yes, there is an extremely short and elegant proof by Carsten Thomassen. See this paper, Prop 2.5.
In fact it is so short that I'll give it in full:
Proof (by induction on $|V(G)|$). If $|V(G)|\le 4$ this holds by inspection so assume $|V(G)|\ge 5$.
Using induction, we can reduce the statement to the case where $G$ is 3-connected. So assume that $G$ is 3-connected. We claim that $G$ has a vertex $v$ which is not contained in any separating triangle. For, if $xyzx$ is a separating triangle in G such that $G - (x, y, z)$ has a component of smallest possible order, then no vertex in that component is contained in a separating triangle of $G$. Now consider any planar embedding of $G$. By the induction hypothesis, the edges of $G - v$ can be coloured in two colours such that no monochromatic triangle occurs. Now we colour the edges incident with $v$ in the same two colours such that no two consecutive edges are part of a monochromatic facial triangle. Then there is no monochromatic triangle at all and the proof is complete.
Added: Gordon has kindly pointed out that Carsten is colouring edges rather than vertices. So the original question is not yet answered.