Does every shape of "lower-bounded width" contain a shape of constant width? Suppose we have a convex set in $\mathbb{R}^n$ whose width along every axis is at least 1. Can we take a subset of its points which has width exactly 1 on every axis?
For example, the unit square contains a Reuleaux triangle:

I suspect the answer here is no, but I haven't come up with any obvious counterexamples. It's also possible that the answer is yes when $n=2$, but counterexamples can be found when $n=3$ (or possibly only at higher $n$).
 A: Aha, solved it.
The counterexample turns out to be quite simple: an equilateral triangle! First, a little geometric reasoning should make it clear that the shortest an equilateral triangle of height $1$ gets along any axis is the altitude from a vertex to the midpoint of the opposite side, as shown below:

For this and the other two such axes on the triangle, for our constant-width subset to have diameter $1$ it must contain both the vertex and the point on the opposite face. So the curve of constant width passes through all three vertices of the triangle. But any two of these vertices are too far apart! So no such subset exists.
A: The equilateral triangle is a counter example. The equilateral triangle with height $1$ (i.e., width $1$ in the vertical axis) has side length $2/\sqrt{3}$.

This set has width larger than one along any axis: given any axis, the at least one line along that axis and through one of the vertices passes through the triangle and the segment inside the triangle has length at least one.
The area of this triangle is $1/\sqrt{3} \approx 0.577$.
The Blaschke–Lebesgue theorem states that the Reuleaux triangle has the least area of all sets of constant width. The area of the Reuleax triangle is $\frac{\pi - \sqrt{3}}{2} \approx .705$.
So the area required for a set of constant width is too large to be contained in the equilateral triangle.
