1.k = 2, find the leftmost square S on the desktop. Suppose its color is C[ 1], all squares with color C[2] intersect it, and these squares contain at least one of the two vertices on the right side of S, so that all squares with color C[2] can be nailed up with two nails.

2. If the proposition holds when k = n and k = n+ 1, you can also find the leftmost square S on the desktop. If its color is C[n+ 1], then all the squares except S whose color is C[n+ 1] will be removed, so the remaining K-colored squares can be divided into two categories, one and the other. Another satisfaction: any k squares with different colors have two squares intersecting (otherwise, K+/kloc-0 squares composed of K squares and S do not intersect and contradict each other). The first one can be nailed with two nails, and the second one can be nailed with 2k-2 nails according to the inductive assumption, that is, the color square can be nailed with 2k-2+2 = 2(k+ 1)-2 nails.