That is: any coloring of the plane with four colors has two points at distance 1 from each other. So says a paper just posted by Aubrey de Grey.
The idea: given a set S of points in the plane, its unit distance graph G_S is the graph whose vertices are S and where two points are adjacent if they’re at distance 1 in the plane. If you can find S such that G_S has chromatic number k, then the chromatic number of the plane is at least k. And de Grey finds a set of 1,567 points whose unit distance graph can’t be 4-colored.
It’s known that the chromatic number of the plane is at most 7. Idle question: is there any chance of a “polynomial method”-style proof that there is no subset S of the plane whose unit distance graph has chromatic number 7? Such a graph would have a lot of unit distances, and ruling out lots of repetitions of the same distance is something the polynomial method can in principle do.
Though be warned: as far as I know the polynomial method has generated no improvement so far on older bounds on the unit distance problem (“how many unit distances can there be among pairs drawn from S?”) while it has essentially solved the distinct distance problem (“how few distinct distances can there be among pairs drawn from S?”)