For proving the equivalence of the Theorems of Borsuk-Ulam and Lusternik-Schnirelmann, we need to cover $S^{n-1}$ with closed sets $F_1,\dots,F_{n+1}$ such that none of the $F_i$ contains a pair of antipodal points. I've read that we can do that by taking a regular $n$-simplex inscribed in $S^{n-1}$ and projecting its faces to $S^{n-1}$, but I'm not satisfied by that argument, since it seems quite hard to specify the vertex coordinates of a regular $n$-simplex inside $S^{n-1}$.
Is there a simpler closed cover to make this work, or a not too messy way to make the simplex-argument rigorous?
Edit: Can we set it up this way?
For $k=1,\dots,n$ let $F_k$ be the set of $(x_1,\dots,x_n)\in S^{n-1}$ where at least one $x_i\le 0$ and $x_k\le x_i$ for all $i$. Let $F_{n+1}$ be the set where all $x_i\ge 0$. All of the $F_k$ are closed, since we are just taking unions and intersections of closed half-spaces. Assuming $x,-x\in F_k$ for $k\le n$ we have $x_k=x_i$ for all $i$, so all coordinates of $x$ are equal and less or equal to $0$, but then $-x\in F_k$ can only happen for $x=0$, a contradiction to $x\in S^{n-1}$. Thus $F_k$ does not contain antipodal points for $k\le n$. We easily verify that $F_{n+1}$ doesn't contain antipodal points either.
Consider the hyperplane, $H$, $x_1+x_2+\dots+x_{n+1}=1$ in $\mathbb R^{n+1}$. It is an $n$-dimensional space. Then consider the points in that hyperplane, $$x_1=(1,0,\dots,0), x_2=(0,1,\dots,0),\dots,x_{n+1}=(0,0,\dots,1)$$ Their convex hull is a regular $n$-simplex.
The points are also points on the sphere in $H$ with center $(\frac1{n+1},\frac1{n+1},\dots,\frac1{n+1})$ and radius $\frac{\sqrt{n^2+n}}{n+1}$.
So all you really need to do is find an isometry between $H$ and $\mathbb R^n$ sending the center to $0$, and you've got your regular $n$-simplex.
It's not pretty, but it's not hard.