The original Haussler-Welzl paper from 1987 may not contain the general statement, but by now this is in monographs and textbooks. See for example Nabil H. Mustafa, Sampling in Combinatorial and Geometric Set Systems, vol. 265 of Mathematical Surveys and Monographs. American Mathematical Society,2022.doi:10.1090/surv/265.

One should be careful: The $\epsilon$ in the Rote-Tichy paper is the measure, i.e. the area of the circle (or volume of the ball in $d$ dimensions), whereas the $\epsilon$ in the Question is the distance, i.e. the radius of the circle. Thus, the claim from the Rote-Tichy paper (and from $\epsilon$-net theory) coincides with that from the Answer.

Recently we have redone the classification of the finite subgroups of $O(4)$, see Towards a geometric understanding of the 4-dimensional point groups, Laith Rastanawi und Günter Rote, arxiv.org/abs/2205.04965 (correcting some mistakes in Conway and Smith).

We have in the meantime redone the classification of the finite subgroups of O(4), see Towards a geometric understanding of the 4-dimensional point groups, Laith Rastanawi und Günter Rote, arxiv.org/abs/2205.04965 (correcting some mistakes in [CS]).

Well, if you scale the triangles to be very big, adding the (relatively small) unit disk changes neither the Hausdorff distance nor the barycenters very much. So the counterexample remains valid to show that this map is not Lipschitz-continuous. Unless you add more assumptions, like that the sets have to be uniformly bounded. (It must be interesting to see how their proof goes.)

In the example of the triangles in the other answer, the Hausdorff distance is as small as you want (on the order of ϵ), while the distance between the barycenters of the two triangles can remain constant (1/3 of the horizontal extension (altitude) of the triangles if the triangles are vertically aligned). Thus, there can be no Lipschitz constant at all for this map.

@FabianWirth. This theorem has more general assumptions, but one can apply it to the current problem: E.g. take the metric space to be the set of compact nonempty convex subsets of $R^n$ itself (with Hausdorff metric), and let $F$ be the identity map. Nevertheless I am surprised by the claimed proof: The example of the triangles in the other answer shows clearly that the barycenter does not work.

Yes. As I said at the very beginning of my answer, an irreducible group forces the moment of inertia to be isotropic.

I found that the continuous version is already in Weyl's 1916 paper (Theorem 5, p.319-320) without detailed proof, for the rationally independent case, and also the monographs contain separate chapters about it. Weyl writes (after the related Theorem 6, on p.321): "Es würde keine Schwierigkeiten machen, die möglichen Ausnahmefälle, [...], vollständig durchzudiskutieren." (It would not pose any difficulties to discuss the potential exceptional cases completely.)