Given a countable coloring of the plane, is it always possible to find a monochromatic set of points $\left\{ \left(x,y\right),\left(x+w,y\right),\left(x,y+h\right),\left(x+w,y+h\right)\right\} $ (the corners of a rectangle)?
This is equivalent to CH. Quoting "Problems and Theorems in Classical Set Theory" by Komjath and Totik, chapter 16, Continuum hypothesis:
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. Previous version, with added explanation about Hamel basis: Using
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. 


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