In the proof of
Theorem: Let $U,V \subset \mathbb{S}^n \hspace{0.2 cm} U\simeq V$. If one of these is open so is the other one.
Proof: Let us call the homeomorphism $h$. Let's consider $N\in I(x)$, with $N \simeq \mathbb{D}^n$ and $\partial N \simeq \mathbb{S}^n$, $y = h(x),N' = h(N) $.
Since $\mathbb{S}^n \setminus \partial N'$ has $N \setminus \partial N'$ and $\mathbb{S}^n \setminus N'$ as connected components these are open in $\mathbb{S}^n \setminus \partial N'$. I don't understand why the connected components are also open in $\mathbb{S}^n$, which is used to conclude the proof.
It seems that $N$ is a neigborhood of $x$ which is homeomorphic to $D^n$. In particular $N$ is compact, thus also $N'$ is compact. If you accept that the connected components of $S^n \setminus \partial N'$ are $A = N \setminus \partial N'$ and $B = S^n \setminus N'$, then clearly $B$ is open in $S^n$ since compact subsets of $S^n$ are closed. Moreover, $\partial N'$ is the topological boundary of $N'$ in $S^n$ (since $\partial N$ is the topological boundary of $N$ in $S^n$ and $h$ is a homeomorphism). There interior $\operatorname{int} N'$ of $N'$ in $S^n$ is open on $S^n$ and we have $$\operatorname{int} N' = N' \setminus \partial N' .$$