A group action is free iff the identity is the only element of the group that fixes any element of X, namely, $\mathbb{R}^2$.
Here is my attempt:
At first, I thought about the natural action, the one that sends the points of $\mathbb{R}^2$ around in a hexagon. But then I realized that every action fixes the origin, and a bigger problem was that every point on the x axis gets fixed by the action $s$, the reflection.
I tried solving this problem by partitioning $\mathbb{R}^2$ into twelve regions using 6 lines, sort of like a tally mark, with the x axis as the long line and 5 vertical lines. Then the regions on top of the x axis can be labeled with $\{e$, $r$, ..., $r^5\}$, and the bottom can be labeled with $\{s$, $sr$, ..., $sr^5\}$.
Then because the cardinality of each region is the same, we can imagine each point getting mapped to another region, depending on which group element of the dihedral group acts on it.
Sort of messy, and the details (especially about which regions get the open intervals) are missing, but I think this takes care of the problem.
I want to make sure that this is correct, and also was wondering if there was a better way to approach this problem. Thanks!