 Solution

Expanding the squared Euclidean distances givesFor distinct centroids, chooseA point assigned to $C_1$ has the displayed distance difference at most zero and therefore $w^{\mathsf{T}}x+b\ge 0$; assignment to $C_2$ gives the reverse inequality. The separating hyperplane is the perpendicular bisector of the line segment joining the two centroids. Points tied between them lie on that hyperplane. This is the nearest-centroid decision hyperplane.

The requested nonzero $w$ needs the additional assumption $z_1\ne z_2$. It does not follow from the stated algorithm alone. In one dimension, take $C_1=\{-1,1\}$ and $C_2=\{0\}$, whose centroids are both zero. A tie rule that retains previous assignments can leave this configuration after an iteration. The inequalities at $-1,1$ imply $b\ge |w|$, while the inequality at zero implies $b\le 0$, forcing $w=0$. Thus the unconditional nonzero-separator claim has this degenerate exception.