co.combinatorics - Finding monochromatic rectangles in a countable coloring of $R^{2}$

This is equivalent to CH.

Quoting "Problems and Theorems in Classical Set Theory" by Komjath and Totik, chapter 16, Continuum hypothesis:

CH holds if and only if the plane can be decomposed into countably many parts none containing 4 different points a,b,c,d such that dist(a,b)=dist(c,d)

This is a stronger requirement than your problem, so assuming CH the answer is no. Their solution, assuming CH is false, proves that there's a monochromatic rectangle.

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Previous version, with added explanation about Hamel basis:

Using

CH holds if and only if R can be colored by countably many colors such that the equation x+y=u+v has no solution with different x,y,u,v of the same color.

This gives a negative answer assuming CH. Explanation: consider R as a vector space over Q. Let A be some basis. Take any bijection A -> A + A, where + is disjoint sum. It induces a linear isomorphism f: R -> R * R. (You can think that there's a linear isomorphism between reals and complexes if that helps.) Then, if you were given a monochromatic rectangle a=(x1, y1), b=(x1+x2, y1), c=(x1, y1+y2), d=(x1+x2, y1+y2), certainly a+d=b+c. Using that isomorphism, f(a)+f(d)=f(b)+f(c) gives a monochromatic solution of quoted equation.