Yes, there is an extremely short and elegant proof by Carsten Thomassen. See [this paper][1], 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.

  [1]: http://www.sciencedirect.com/science/article/pii/0095895686900614