Consider a directed graph $G$, each vertex represents a point of $\mathcal{S}$ and we connect two vertices $a \rightarrow b$ if $b$ is the closest point from $a$. Graph $G$ consists of several connected components, in which each component consists of at least two vertices, and every point in a component must be coloured with the same colour. Also, a connected component is a union of a cycle and disjoint paths. Note that no cycle with a length greater than two may occur, since if $p \rightarrow q \rightarrow r$ is a path, then $d(q, r) < d(p, q)$ since $r$ is the closest point from $q$. If a point $V$ with colour $\mathcal{C}$ has its furthest point in a different connected component, there are at least four points with colour $\mathcal{C}$. Otherwise, if its furthest point is in the same connected component, then it's possible that there can only be three points with colour $\mathcal{C}$. Let's say the colour that only has three points as nice colour. We claim that there is only at most one nice colour. For a nice colour, its points must be in the same connected component. Let's say the points and the edges $a \rightarrow b \leftrightarrow c$ (So $b$ is the closest point from $a$, $c$ is the closest point from $b$, and $b$ is the closest point from $c$). We must have $c$ as the furthest point from $a$, and $a$ as the furthest point from both $b$ and $c$. Let say we have another nice colour with connected component $a_1 \rightarrow b_1 \leftrightarrow c_1$. Then since $a$ is the furthest from $b$ and $b_1$ is the closest from $a_1$, we have: \[ d(a, b) > d(a_1, b) > d(a_1, b_1), \]and similarly by symmetry we should have $d(a_1, b_1) > d(a,b)$. A contradiction. Therefore, there are at most one nice colour. If $k$ denotes the number of the colours, we have: \[ 2023 \ge 4(k-1)+3 \implies 506 \ge k. \]

Put three points $A(-1,0), B(1,0), C(2,0)$ and construct two circles with center $A$ and $B$, each with radius $2$. We colour $A, B,$ and $C$ with the same colour. Consider a circle $\Gamma$ with center on the $x$-axis and on the line connecting the centers of the two small circles, that intersect them (the small circles are symmetric w.r.t. $x$-axis). Pair the remaining $2020$ points, there are $1010$ such pairs, and put half of the pairs to the arc of $\Gamma$ inside the top small circle, and the rest to the arc of $\Gamma$ inside the bottom one (each pair is diametrically opposite to another pair in the top circle, so they are the furthest from that pair). Points in each pair lie very close to each other, so they are the closest to each other. Each pair of points in the top small circle will correspond to another pair of points in the bottom circle as their furthest points, and the four points from these two pairs will have the same colour, so there are $1010/2 + 1 = 506$ colours. [asy][asy]import graph;size(8cm); pair A=(-1,0),B=(1,0),D=(0,-sqrt(3)), C=(2,0); real t = 0.3, r = 0.1; pair e1 = (t, sqrt(3)), e2 = (t, -sqrt(3)); filldraw(D--arc(B,D,(0,sqrt(3)))--arc(A,(0,sqrt(3)),D,direction=CW)--cycle,0.3*cyan+0.7*white); draw(circle(A,2)^^circle(B,2)); draw(circle(e1,r)^^circle(e2,r)); filldraw(circle(e1,r)^^circle(e2,r), magenta); draw(circle((t,0), sqrt(3)), dashed); dot(A^^B^^C); MP("A",A,SW); MP("B",B,SE); MP("C",C,SE); MP("\Gamma",(2,0.8), N); [/asy][/asy]

This is a cute problem. I misread the problem while first trying to solve this, not seeing that $P$ must also have the same colour as the closest and furthest point from it (I thought only the two points, the furthest and closest from $P$, have the same colour, and $P$ can be different). I think this variant is harder (?)

Note that no cycle with a length greater than two may occur, since if $p \rightarrow q \rightarrow r$ is a path, then $d(q, r) < d(p, q)$ since $r$ is the closest point from $q$.

Note that no cycle with a length greater than two may occur, since if $p \rightarrow q \rightarrow r$ is a path, then $d(q, r) < d(p, q)$ since $r$ is the closest point from $q$.