Suppose $U,V \subset \mathbb R^n$ are open subsets such that $U \subset V$. Let $x \in \partial U \cap \partial V$, and suppose there is $\rho>0$ such that $$ B_{\rho}(x) \cap \partial U = B_{\rho}(x) \cap \partial V, $$ where $B_{\rho}(x)$ is the ball of radius $\rho$ centered at $x$ in $\mathbb R^n$.
Question: Is it true that $$B_{\rho} \cap U = B_{\rho} \cap V\,?$$
Some thoughts: My initial attempt was to prove this is true, using a connectedness argument. If $C \subset B_{\rho}(x) \cap V$ is a connected component, then by path-connectnedness it follows that either $C \subset U$ or $C \subset V \setminus \overline U$. Also $\partial C \subset \partial (B_{\rho}(x) \cap V) \subset (\partial B_{\rho}(x)\cap V) \cup (B_{\rho}(x) \cap \partial V)$. However I don't see how to rule out the possibility that $C$ is a non-empty subset of $V \setminus \overline U$. I'm also starting to question whether this is even true, since boundaries of general open sets can ill-behaved in general.
The answer is no, for any $n \ge 1$.
For $n = 1$, take $U = (-\infty, 0)$ and $V = (-\infty, 0) \cup (0, \infty)$. Note that $U$ is a proper subset of $V$ and they have the same boundary, namely $\{0\}$. Now it's easy to take $x = 0$ and $\rho = 1$, and we see that for $B = B_\rho(x)$, we have $B \cap \partial U = B \cap \partial V$ but $B \cap U \neq B \cap V$.
For $n > 1$, you can take $U = (-\infty, 0) \times {\Bbb R}^{n - 1}$ and $V = ((-\infty, 0) \cup (0, \infty)) \times {\Bbb R}^{n - 1}$, similarly.
(A somewhat more interesting example: take $A$ to be the set of points with at least one zero coordinate, and look at the $2^n$ connected components of the complement of $A$. These will be the open quadrants or octants or whatever. Take $V$ to be the union of all those components, and $U$ to be the union of all except one.)
Probably the next question is, what if we assume $U$ and $V$ are connected?