I've came across this proof (anwer of Neil Strickland to a MathOverflow question) of the fact that $\mathbb{R}P^{n-1}$ cannot be embedded in $\mathbb{R}^n$ for $n > 2$. There are a few details that I don't understand and would really appreciate if someone further elucidated them to me.
I understand that if an embedding $\mathbb{R}P^{n-1} \subseteq S^n$ existed, then $S^n \setminus \mathbb{R}P^{n-1}$ would have two connected components. I don't know why their closures, $A$ and $B$, have the claimed properties $A\cap B = \mathbb{R}P^{n-1}$, $A \cup B = S^n$. Furthermore, the version of the Mayer-Vietoris theorem I am aware of requires the union of interiors of $A$ and $B$ to cover $S^n$.
Why does the isomorphism $H^n(A) \times H^n(B) \cong H^n(S^n\setminus \mathbb{R}P^{n-1})$ hold? This is clear to me if $A$ and $B$ are replaced by the respective connected compontents.
One of the comments to Neil Strickland's answer brings attention to the fact that his proof is very similar to the one of Thom (Theorem V.15 in "Espaces fibrés en sphères et carrés de Steenrod" numdam.org/item?id=ASENS_1952_3_69__109_0). I tried looking at that paper but I didn't achieve much, as I don't speak French. Google Translate seems to imply, however, that it is assumed that $\mathbb{R}P^{n-1}$ is embedded nicely enough, so that the closures of the connected components indeed have the properties claimed in 1. Is this in fact the case?