The generalized Lusternik-Schnirelmann Theorem states that
If $S^n$ is covered by $n+1$ sets $A_1, A_2, ... ,A_{n+1}$ such that each $A_i$ is either open or closed, then there exists an $i$ such that $A_i$ has a pair of antipodal points.
I'm having trouble proving this theorem for "$A_i$ is either open or closed". I can prove the theorem given the hypothesis that all $A_i$ are closed.
Define $d_i : S^n \to \mathbb{R}$ by $d_i(x) = \inf \{ |x - y| : y \in A_i \}$. Clearly, this function is continuous. Now, consider the map \begin{align*} \Psi: S^n & \longrightarrow \mathbb{R}^n \\ x & \longrightarrow (d_1(x), ..., d_n(x)). \end{align*} From the Borsuk-Ulam theorem, there exists an $x \in S^n$ such that $\Psi(x) = \Psi(-x) = \Omega$. If any of the coordinates of $\Omega$ are $0$, then, since the $A_i$ are closed, $x, -x$ must be limit points of some $A_i$ and hence are in that $A_i$. If none of the coordinates are $0$, then $x, -x$ are both in $A_{n+1}$.
If the $A_i$'s could also be open, the above argument does not work.
Any suggestions on how to prove the theorem would be appreciated.
You can extend your argument as follows: let $x$ be defined as above. If $x,-x\in A_{n+1}$ we're done, so WLOG assume $x\not\in A_{n+1}\implies x\in A_i$ for some $i\neq n+1$. Thus the $i$th coordinate of $\Omega$ is $0$, so if $A_i$ is closed we're done by your proof. If $A_i$ is open, then $-x\in \bar A_i$ so $B(x,\delta)\subseteq A_i$ for some $\delta>0$ and $B(-x,\delta)\cap A_i\neq\emptyset$, so if we pick some $y$ from the latter then $y,-y\in A_i$.