I have a feeling that the following statement is true (it's not from any book or paper, but it seems plausible):
If $U\neq \emptyset$ is open and precompact (that is, $\overline{U}$ is compact) in $\mathbb{R}^n$, then the boundary of $U$ has a subset $F\subset \partial U$ homeomorphic to an open set of $\mathbb{R}^{n-1}$, or equivalently to $\mathbb{R}^{n-1}$ itself.
The reason I think this is true is it is pretty easily shown for $n=1$, since the boundary of such an open set consists of isolated points. And also, if you think about the general case you have that this is true for open balls in $\mathbb{R}^n$, since the boundary is homeomorphic to $S^{n-1}$.
But I get stuck when I try to think of a way to solve this problem for any open precompact subset.
Notice that I'm not trying to show that $\partial U$ is locally euclidean, since there is a simple counter-example in $\mathbb{R}^2$, where on can get an X-shaped boundary for certain open precompact subsets. I only want to show that some subset of the boundary is locally euclidean (actually euclidean as whole).
I am open for suggestions on how to prove/disprove this.
Thank you all!
The answer is negative. I don't know a simpler example, but here is one using ideas due to Bing.
In this paper (Example 2), Bing proves that there is a hereditarily indecomposable continuum which separates the plane. I will break down this notion and explain why it finishes the problem.
A continuum is a nonempty, compact, connected metric space (here we will only use subsets of the plane, no need to worry about abstract metric spaces). An example of a continuum is the interval $[0,1]$.
A continuum is indecomposable if it is not a union of two proper subcontinua (=subsets each of which is a continuum as well). $[0,1]$ is decomposable, since it is a union of $[0,1/2]$ and $[1/2,1]$.
A continuum is hereditarily indecomposable if each of its subcontinua is indecomposable as well. Since $[0,1]$ is decomposable, a hereditarily indecomposable continuum cannot contain a copy of an interval, and hence cannot contain a copy of $\mathbb R^1$.
Now, it's far from clear that hereditrarily indecomposable continua (with more than one point) exist. The first example was a pseudo-arc, constructed by Knaster. Bing's example I refer to is really just a variant of Knaster's construction.
When I say that this continuum $C$ separates the plane I mean that the complement $\mathbb R^2\setminus C$ has more than one connected component. It's easy to see (by compactness of $C$ that there can be only one connected component which is unbounded, so let $U$ be any bounded component of $\mathbb R^2\setminus C$. Now it's clear that $\partial U$ is contained in $C$, so by what was said above, it contains no subset homeomorphic to $\mathbb R^1$.