The question is in the title.
The cited proof is more or less as following: taken $U\subseteq \mathbb{R}^n$, $h\colon U \rightarrow \mathbb{R}^n$ an embedding, he picks $x_1\in h(U)$ then considers $x=h^{-1}(x_1)$, a ball $B=B(x,\varepsilon)$ such that $\bar{B}\subseteq U$. Now $\mathbb{R}^n\setminus h(\bar{B})$ and $h(\bar{B}\setminus Fr[B])$ are nonempty, connected and disjoint, so they are the components of $\mathbb{R}^n\setminus h(Fr[B])=(\mathbb{R}^n\setminus h(\bar{B}))\cup h(\bar{B}\setminus Fr[B])$. Then he infer that these sets are open and he concludes the proof.
What isn't clear to me is why the two sets are the components, I think he should know that both sets are open (or closed) because for example $[0,1]=[0,1)\cup \{1\}$ and both $[0,1)$ and $\{1\}$ are nonempty, connected and thay are disjoint but $[0,1]$ is connected.
Thanks to the DanielFischer's comment I think I have solved the question.
Yes, in the previous paraghraph there is the proof that any embedding of $S^{n-1}$ in $\mathbb{R}^n$ disconnects the space (it is not proven that the components are two, but in this case is not relevant). So $h(Fr[B])$ is the image of an embedding of $S^{n-1}$ in $\mathbb{R}^n$, so its complement is disconnected, so the reasoning is corrected.