Nice problem! We will first dispatch of the case where there are two points which are at least $\sqrt3$ apart. Let $S$ denote the set of points in question. Case 1. The diameter of $S$ is $\ge \sqrt 3.$ Let $A, B$ be points in $S$ so that $AB \ge \sqrt3.$ Let $\omega_A, \omega_B$ denote the unit circles centered at $A, B,$ respectively. Let $\omega$ denote the region which is the (closed) intersection of $\omega_A$ and $\omega_B.$ Note that since $AB \ge \sqrt 3$, $\omega$ is either empty or has diameter at most $1.$ We claim that all of the points of $S$ in $\omega_A / \omega$ comprise a set of diameter at most $1.$ Indeed, if not, then consider two points $P_1, P_2 \in (\omega_A / \omega) \cap S$ so that $P_1 P_2 > 1.$ Then $P_1, P_2, B$ contradict the condition of the problem. A similar thing holds for $S \cap (\omega_B / \omega),$ and so from the previous results we know that $S \cap (\omega_A / \omega), S \cap (\omega_B / \omega), S \cap \omega$ is a partition of $S$ into three parts, each of diameter at most $1.$ Case 2. The diameter of $S$ is at most $\sqrt 3.$ Notice that for any three points $A, B, C$ so that $AB, BC, CA \le \sqrt 3$, there is a unit circle covering $\triangle ABC.$ Indeed, simply consider cases on whether $\triangle ABC$ is acute or obtuse (if acute, take the circumcircle; if obtuse take the circle with the longest side of $\triangle ABC$ as diameter). Hence, by Helly's Theorem, the set of unit circles centered at the points in $S$ all have a common point, say $P$. This means that the entirety of $S$ is contained in the unit circle centered at $P.$ Let $Q \in S$ be the point furthest from $P$, and let $PQ = r \le 1.$ Let $\omega_P$ denote the circle centered at $P$ of radius $r$; note that $S \subset \omega_P.$ Let $\omega_Q$ denote the unit circle centered at $Q.$ Note that the set of points of $S$ contained in $\omega_P / \omega_Q$ has diameter at most one, as else we can obtain a contradiction with two points of distance $>1$ and $Q.$ It only remains to notice that $\omega_P \cap \omega_Q$ can be split into two sets of diameter $1$ rather easily, by simply bisecting it along line $PQ.$ From here, the partition of $S$ is clear. As we've exhausted all cases, we're done. $\square$