Let some vertex of a cube be $A$ and let $B$, $C$ and $D$ be the opposite vertices of the faces of the cube that $A$ belongs to (see fig. 9). Then of the vertices $B$, $C$ and $D$ any two are also the opposite vertices of some face of the cube. Therefore any two of the chosen four vertices are at the distance of a face diagonal of the cube. Therefore any three form an equilateral rather than a right triangle.

Let us now look at the situation where we choose at least 5 vertices. Two opposite faces of the cube include all the vertices of the cube. Therefore at least one of the two opposite faces has to include at least 3 of the chosen vertices. But three vertices of a square form a right triangle.