Is it possible to color each point of the plane red or blue so that no square with unit side length and monochromatic vertices is formed?
EDIT: I think this is possible,for each point in the plane color the points that are unit length apart with an opposite color.This way it is guaranteed that no monochromatic vertices of unit length apart is formed and so no square with unit side length and monochromatic vertices is formed.
EDIT: The point I was trying to make was that there must be at least one point in the unit length square with a different color, if not it violates the coloring defined above. Hence, it is possible to color points such that no square with unit side length and monochromatic vertices is formed.
Best Answer
Horizontal bands with height exactly $1$, closed on the bottom, open on the top, provide such a coloring. Thus, $(x, y)$ is red if $2|\lfloor y \rfloor$ and blue otherwise.
Number the square's vertices in order of height and let $\theta$ be the angle between a horizontal line passing through vertex $1$ and the side of the square $\overline{12}$. Without loss of generality, $0 \leq \theta \leq \frac{\pi}{2}$. (Otherwise perform a reflection.) Let $y_1$ be the $y$-coordinate of vertex $1$. Then the $y$-coordinates of the four vertices are:
$$y_1\\ y_2 = y_1+\sin \theta\\ y_3=y_1+ \cos \theta \\ y_4=y_1+\sin \theta + \cos \theta = y_1+\sqrt{2}\sin (\theta + \frac{\pi}{4}).$$
The maximum height difference between any two vertices is $\sqrt{2} \lt 2$ so a monochromatic square is possible only if the height difference between some two consecutive vertices is strictly greater than $1$.
But that can't possibly happen. It obviously can't happen between $1$ and $2$, $3$ and $4$, $2$ and $4$, or $3$ and $1$. And $y_3-y_2= \cos \theta - \sin \theta = \sqrt{2} \cos (\theta + \frac{\pi}{4})$ so $0 \leq \theta \leq \frac{\pi}{2} \Rightarrow |y_3-y_2| \leq 1$, completing the proof.
Edited to respond to comments requesting additional detail.