Update: The argument I previously provided was incorrect. I will now outline a correct argument, followed by the incorrect argument and an explanation of what went wrong.
Correct Explanation:
By Alexander duality, we can write decompositions into exactly two connected components, for each of the following two sets:
$$S^n-h\big(S^{n-1}\times (0,1)\big)=C\sqcup D$$
$$S^n-h\big(S^{n-1}\times [0,1]\big)=C'\sqcup D'$$
We set our naming convention so that $C=C'\sqcup S^{n-1}_1$ and $D=D'\sqcup S^{n-1}_0\!.$ To see that this makes sense, we must argue that the four connected sets $C'\!,\, D'\!,\, S^{n-1}_0$ and $S^{n-1}_1$ (whose disjoint union is precisely $C\sqcup D$) are reasonably distributed between $C$ and $D$ :
- Suppose $S^{n-1}_0\sqcup S^{n-1}_1\subseteq C.$ Then we can write $$S^n=C\sqcup h\big(S^{n-1}\times (0,1)\big)\sqcup D=\Big(C\cup h\big(S^{n-1}\times [0,1]\big)\Big)\sqcup D.$$
Because $C$ and $D$ are closed, this contradicts the connectedness of $S^n\!.$ Analogously, we cannot have $S^{n-1}_0\sqcup S^{n-1}_1\subseteq D,$ so we can assume that $S^{n-1}_1\subseteq C$ and $S^{n-1}_0\subseteq D$.
- Suppose $C'\!\sqcup D'\subseteq C$. Since $C'$ and $D'$ are open, we can see that $$C-D'=C'\sqcup S^{n-1}_1$$ is a closed set containing $C'\!,$ which is not itself closed. Thus $\overline{C'}$ intersects $S^{n-1}_1$ nontrivially. We similarly conclude that $\overline{D'}$ intersects $S^{n-1}_1$ nontrivially. Also note that $$\overline{C'}\cap S^{n-1}_0\subseteq C\cap D=\emptyset$$ and similarly $\overline{D'}\cap S^{n-1}_0=\emptyset.$ Thus $$S^n-S^{n-1}_0=\overline{C'}\cup \overline{D'}\cup h\big(S^{n-1}\times (0,1]\big)$$ is connected, which contradicts the Jordan-Brouwer separation theorem. Analogously, we cannot have $C'\!\sqcup D'\subseteq D,$ so we can assume that $C'\subseteq C$ and $D'\subseteq D$.
Since $C'$ is open and $C=C'\sqcup S^{n-1}_1$ is closed, we can see that $\overline{C'}$ intersects $S^{n-1}_1$ nontrivially, so $\overline{C'}\cup h\big(S^{n-1}\times(0,1]\big)$ is connected. Hence, we have a decomposition $$S^n-S^{n-1}_0=\Big(\overline{C'}\cup h\big(S^{n-1}\times (0,1]\big)\Big)\sqcup D'$$
into two connected sets, which therefore must be the connected components of $S^n-S^{n-1}_0\!.\,$ This implies that $D'=B$, since the other component contains $S^{n-1}_1\!.\,$ Analogously, we can conclude that $C'=A.$ Then it is clear that
$$S^n=C'\sqcup D'\sqcup h\big(S^{n-1}\times [0,1]\big)=A\sqcup B\sqcup h\big(S^{n-1}\times [0,1]\big)$$
Incorrect Explanation:
Let $q:S^n\rightarrow X$ be the quotient map formed by collapsing $\overline A$ and $\overline B$ to points $a$ and $b$.
It is not too hard to show that $X$ is Hausdorff, so any injective map from a compact space into $X$ is an embedding. Observe that $X$ is locally Euclidean, because:
- $X-\{a,b\}\cong S^n-(\overline A\cup \overline B)$ is an open submanifold of $S^n$;
- The restriction $h:S^{n-1}\times [1/2,1]\rightarrow S^n$ descends to an embedding $D^n\rightarrow X$ of a closed $n$-cell containing $a$ in its interior (we view $D^n$ as coming from $S^{n-1}\times [1/2,1]$ by collapsing $S^{n-1}\times 1$);
- The restriction $h:S^{n-1}\times [0,1/2]\rightarrow S^n$ descends to an embedding $D^n\rightarrow X$ of a closed $n$-cell containing $b$ in its interior (we view $D^n$ as coming from $S^{n-1}\times [0,1/2]$ by collapsing $S^{n-1}\times 0$).
The map $h:S^{n-1}\times [0,1]\rightarrow S^n$ similarly descends to an embedding $p:S^n\rightarrow X$ (where we view $S^n$ as coming from $S^{n-1}\times [0,1]$ by collapsing $S^{n-1}\times 0$ and $S^{n-1}\times 1$ to points). This embedding $p$ must be a homeomorphism, since $S^n$ is compact and $X$ is connected (and both are topological $n$-manifolds). The fact that $p$ is surjective is equivalent to the fact that I had asked about, namely that $$S^n=A\cup B\cup h(S^{n-1}\times [0,1]).$$ The quotient map that I mentioned in my question is then simply $p^{-1}\circ q$.
What went wrong?
When I tried to define a chart about $a$ (and similarly for $b$, but I'll stick with one point for clarity), I described an embedding $D^n\rightarrow X$ containing $a$ in the image of the disk's interior, but I actually needed to argue that this image contained a whole neighborhood of $a$. My conclusion that $p$ was surjective came from a corollary to "invariance of domain," the proof of which relies crucially on the openness of charts in a topological manifold (although the proof does not require the Hausdorff or second-countable assumptions). For the "chart" that I defined to map onto an open set, we need $A\cup h\big(S^{n-1}\times (1/2,1]\big)$ to be open in $S^n\!.\,$ But if we could show this, then we could argue directly that $A\cup h\big(S^{n-1}\times (0,1]\big)$ is clopen and connected in $S^n-S^{n-1}_0\!,\,$ which means that it must be the other connected component (besides $B$). Then we get
$$S^n-S^{n-1}_0=A\cup h\big(S^{n-1}\times (0,1]\big)\cup B,$$ which clearly implies the desired result, without having to worry about $X$ being locally Euclidean.
