6. Let $S$ be the set of points inside and on the boundary of a regular hexagon with side length 1. Find the smallest $r$ such that there exists a way to three-color the points in $S$ so that the distance between any two points of the same color is less than $r$.
6. $r_{\min }=\frac{3}{2}$.

On one hand, construct the following coloring: let the regular hexagon $A B C D E F$, with center $O$. Draw perpendiculars from $O$ to $A F$, $B C$, and $D E$, with feet at $X$, $Y$, and $Z$ respectively. Color the three regions marked in the diagram 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 the regions can be colored with the color of either adjacent 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 this 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 $\frac{3}{2}$, meaning $r=\frac{3}{2}$ is feasible.
Next, we prove that when $r<\frac{3}{2}$, a contradiction arises. Consider a circle of radius $r$ centered at a point $A$. For any point $B$ on the circle, there exists a positive $\varepsilon$ such that if $A^{\prime}$ is another point on the circle satisfying $A A^{\prime}<\varepsilon$, then $A B>r$, $A^{\prime} B>r$, and $A A^{\prime}<\varepsilon$. Since $A$ and $B$ are colored differently, and $A^{\prime}$ and $B$ are also colored differently, $A^{\prime}$ must be colored the same as $A$. Therefore, any two points within a distance of $\varepsilon$ must be colored the same, leading to the entire circle being colored the same, which is a contradiction! Thus, $r$ must not be less than $\frac{3}{2}$.
In conclusion, $r_{\min }=\frac{3}{2}$.