Let us say that a set of points in $\mathbb{R}^d$ is minimal if it forms exactly the set of vertices of a convex polytope. Equivalently, no proper subset of the points has the same convex hull; no point is contained in the convex hull of the others.
Then there is a well-known application of Ramsey's theorem to discrete geometry that says the following:
(1) For any integer $m$, there exists an integer $n$ such that any $n$ points in $\mathbb{R}^2$, no three collinear, contains a minimal subset of $m$ points.
In this case, since it's $\mathbb{R}^2$, a minimal set is just a convex $m$-gon. You can find the proof here.
The question is simply replacing $\mathbb{R}^2$ with $\mathbb{R}^d$ above.
Prove or disprove: (2) For any integer $m$, there exists an integer $n$ such that any $n$ points in $\mathbb{R}^d$, no three collinear, contains a minimal subset of $m$ points.
The proof of (1) relies on two facts: (i) that a set of points in $\mathbb{R}^2$ is minimal if and only if every subset of 4 points in it is minimal, and (ii) that any set of five points contains a minimal subset of 4 points. (ii) fails in $\mathbb{R}^d$. So I am not sure the argument given at the cut-the-knot link extends in any obvious way. (An almost identical argument is given in Douglas B West's Introduction to Graph Theory, 2nd edition, page 382.)
I believe that the argument in the cut-the-knot link does extend to $\mathbb{R}^d$ (at least, it extends to $\mathbb{R}^3$, with some work) if in (2) the $n$ points are in general position. But we only said no three were collinear.