CLRS closest-pair $L_m$ distances
One approach to see what is going on here is via the notion of an open ball under a given distance. Given a distance $d$, the open ball of radius $r$ around a point $p$ is given by $B_r(p):=\{q\mid d(p,q)&lt;r\}$, i.e. the set of all points with distance less than $r$ to the center of the ball, $p$. In the plane and under the Euclidean distance, a ball is an open disk (a disk with the boundary removed) of radius $r$. 

Since $\delta$ is the minimum distance between any pair of points in the left half, any pair of points in the left $\delta\times\delta$ square has a distance of at least $\delta$. This is equivalent to saying that the open ball of radius $r$ centered at a point $p\in P_L$ contains no other point $q\in P_L$. So, the maximum number of points from $P_L$ that can be in the $\delta\times \delta$ square is the largest number of points you can 'fit' into the square such that the ball surrounding each of those points contains no other point from $P_L$.

An open ball under the Euclidean distance is an open disk. If you place four points at the corners of the $\delta\times \delta$ square, the open disks of radius $\delta$ centered at those points together cover the entire square, so there is no position in the square to place for a fifth point.

However, under the Manhattan distance, an open ball of radius $\delta$ is an open square with diagonal length $\delta$, rotated by $45$ degrees (an "open diamond"). If we place such an open diamond at each corner of the $\delta\times \delta$ square, then there is exactly one position in the square not covered by an open diamond: the center of the square. This is the location where you can place a fifth point under the Manhattan distance.



As for the $L_m$-distance with $m&gt;2$, note that the balls under these metrics all strictly contain the ball under the $L_2$-distance. I'll leave it up to you to use this fact to determine the maximum amount of points that can be in the $\delta\times\delta$-square.