To ensure there exists a way to three-color the points in $S$, such that after coloring, the distance between any two points of the same color is less than $r$.

Make sure there is a method to three-color the points in $S$, so that after coloring, the distance between any two points of the same color is less than $r$.
Parse 1 $r_{\min }=\frac{3}{2}$.
On one hand, construct the following coloring (as shown in Figure a): Let the center of the regular hexagon $A B C D E F$ be $O$, and draw perpendiculars from $O$ to $A F$, $B C$, and $D E$, with the feet of the perpendiculars being $X$, $Y$, and $Z$ respectively. Color the three regions marked in the figure with colors I, II, and III respectively. Specifically, points $X$, $Y$, and $Z$ are colored with colors I, II, and III respectively (the points at the intersections of different regions can be colored with the color of either region). In this way, the maximum distance between any two points in the same region is $X Y=Y Z=Z X=\frac{3}{2}$, and the maximum value is not achieved by two points in the same region. Therefore, the distance between any two points of the same color is less than $\frac{3}{2}$, i.e., $r=\frac{3}{2}$ is feasible.

Now consider the points on the boundary of the hexagon. For any two points $A^{\prime}$ and $B^{\prime}$ on the boundary, the minimum length required to travel from $A^{\prime}$ to $B^{\prime}$ along the boundary of the hexagon is called the boundary distance between $A^{\prime}$ and $B^{\prime}$ (for example, the boundary distance between $A$ and $E$ is $A F + F E = 2$). It can be proven: for $01$, so $M$ and $N$ cannot be on the same side, and we consider the following three cases:
(1) If $M$ and $N$ are on adjacent sides (as shown in Figure b), then
$$
\begin{array}{c}
A M + A N \geqslant 2 - \varepsilon, \\
M N = \sqrt{A M^{2} + A N^{2} + A M \cdot A N} \geqslant \frac{\sqrt{3}}{2}(A M + A N) \\
\geqslant \frac{\sqrt{3}}{2}(2 - \varepsilon) = \sqrt{3} - \frac{\sqrt{3}}{2} \varepsilon > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon.
\end{array}
$$
(2) If $M$ and $N$ are on sides separated by one side (as shown in Figure c), then
$$
A M + A B + B N \geqslant 2 - \varepsilon, \text{ i.e., } A M + B N \geqslant 1 - \varepsilon.
$$

Draw perpendiculars from $M$ and $N$ to $A B$, with the feet of the perpendiculars being $M_{1}$ and $N_{1}$, then
$$
\begin{aligned}
M N & \geqslant M_{1} N_{1} = M_{1} A + A B + B N_{1} = \frac{A M}{2} + A B + \frac{B N}{2} \\
& \geqslant 1 + \frac{1 - \varepsilon}{2} = \frac{3}{2} - \frac{\varepsilon}{2} > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon.
\end{aligned}
$$
(3) If $M$ and $N$ are on opposite sides, then
$$
M N \geqslant \sqrt{3} > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon.
$$

In summary (1)-(3), there exist two points $M$ and $N$ such that $M N > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon$, i.e., $r > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon \Rightarrow \lim _{\epsilon \rightarrow 0+} r \geqslant \lim _{\epsilon \rightarrow 0+}\left(\frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon\right)$, i.e., $r \geqslant \frac{3}{2}$.
In summary, $r_{\min }=\frac{3}{2}$.
Parse 2 $r=\frac{3}{2}$ is feasible, see method one. Now prove that $r0$, such that $r+\varepsilonr, A^{\prime} C>r$, but $B$ and $C$ are colored differently, so $A^{\prime}$ must be the same color as $A$. Therefore, for any two points with a distance less than $\varepsilon$, they must be the same color, thus the entire circle will be one color, which is a contradiction.