Distribution of points on a rectangle

The problem and solution are in Jiří Herman, Radan Kučera, Jaromír Šimša, Counting and Configurations: Problems in Combinatorics, Arithmetic, and Geometry, page 272. Let the rectangle have corners $(0,0),(0,3),(4,0),(4,3)$. Draw line segments joining $(0,2)$ to $(1,1)$ to $(2,2)$ to $(3,1)$ to $(4,2)$, also $(1,1)$ to $(1,0)$, $(2,2)$ to $(2,3)$, and $(3,1)$ to $(3,0)$. This splits the rectangle into $5$ pieces, and it's not hard to show two points in the same piece must be within $\sqrt5$.

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A picture to illustrate the solution.

_{For a different approach, also start by observing that any rectangular area with sides 2 and 1 can contain at most one point if any two points are to be more than $\sqrt{5}$ apart. Now note that 2-by-1 rectangles can be placed horizontally (blue in illustration below) or vertically (green). So we can...}

Split the 4-by-3 rectangle into 12 unit size squares, color like a chessboard (black and white such that no squares sharing an edge have the same color). Consider a pair of neighboring squares: If they contain (at least) two points, these have a distance $\leq \sqrt{5}$. If they contain no point, there must be another pair of adjacent squares with two points (pigeonhole principle, five 2-by-1 rectangles/square pairs left for six points), whose distance is then $\leq \sqrt{5}$. If each black/white pair of adjacent squares contains exactly one point, it follows that the points must be either all on black or all on white squares (there are exactly 6 of each).

Then, in either case there must be a "diagonal" of three squares with one point each (***** or x in illustration). Now imagine the middle square split along the other diagonal. Now for each triangle, every point in or on it is no farther than $\sqrt{5}$ from any point in the respective closer, diagonally "neighboring" square! (In other words, w.r.t. to the illustration, if the point in the middle square is closer to the top left, it is less than $\sqrt{5}$ from the point in the top left, otherwise it is within $\sqrt{5}$ distance of the point in the lower right square.)