If $U\subseteq V\subseteq \mathbb{R}^{2}$ then $\partial V=(\partial (V\setminus U) \cup \partial U)\cap \partial V$? Let $U\subseteq V\subseteq \mathbb{R}^{2}$ be two sets. Does it hold that $\partial V=(\partial (V\setminus U) \cup \partial U)\cap \partial V$? I.e. $\partial V \subseteq (\partial (V\setminus U) \cup \partial U)$? If not, what additional properties are needed for $U$ and $V$?
 A: Let $x\in\partial V$.  If $x\not\in\overline{U}$, let $W$ be an open set containing $x$.  Now consider $W\setminus\overline{U}$, this is an open set containing $x$.  Moreover, since $x\in\partial V$, there is some point of $V$ and a point not of $V$ in this open set.  But since we're avoiding $\overline{U}$, it must be that this open set contains a point of $V\setminus U$ and a point outside of $V$ (so outside of $V\setminus U$).  So $x\in\partial(V\setminus U)$.
If $x\in\overline{U}$, then $x$ must be in the boundary of $U$, otherwise, there would be an open set $W'$ containing $x$ entirely contained within $U$, but then $W'$ would be entirely contained within $V$, so $x$ would not be a boundary point.  So $x\in\partial U$.
A: For any subset $A$ of a topological space, $$\partial A\equiv\operatorname{cl} A\setminus(\operatorname{int} A)=\operatorname{cl} A\cap\operatorname{cl}A^{\mathsf c}.$$
In the particular question, it is sufficient to prove the following

Claim: If $U\subseteq V\subseteq\mathbb R^2$, then $$\partial V\subseteq \partial (V\setminus U)\cup\partial U.$$

Proof: Suppose that $x\in\partial V=\operatorname{cl} V\cap\operatorname{cl}V^{\mathsf c}$. The goal is the show that either $x\in\partial(V\setminus U)$ holds or $x\in\partial U$ holds. If $x\in\partial U$, then we’re done. If $x\notin\partial U=\operatorname{cl}U\cap\operatorname{cl}U^{\mathsf c}$, then either $x\notin\operatorname{cl}U$ or $x\notin \operatorname{cl} U^{\mathsf c}$ must hold. We must show that in either of the two cases, $x\in\partial (V\setminus U)$ is true, which will complete the proof.
Case 1: $x\notin \operatorname{cl} U$. Let $F\subseteq\mathbb R^2$ be any closed set such that $$V\cap U^{\mathsf c}\subseteq F\tag{$\spadesuit$}.$$ It is clear that $$V\cap U\subseteq \operatorname{cl}V\cap \operatorname{cl} U\tag{$\heartsuit$}.$$ Therefore, taking unions of both sides of ($\spadesuit$) and ($\heartsuit$), one has that $$V=(V\cap U^{\mathsf c})\cup (V\cap U)\subseteq F\cup(\operatorname{cl}V\cap\operatorname{cl}U).$$ Hence, $F\cup(\operatorname{cl}V\cap\operatorname{cl}U)$ is a closed set containing $V$, and since $x\in\partial V\subseteq\operatorname{cl} V$, it must be the case that $$x\in F\cup(\operatorname{cl}V\cap\operatorname{cl}U).$$ But $x\notin\operatorname{cl} U$ by the leading assumption for this case, so it follows that $x\in F$. We have just proven that if $F$ is a closed set containing $V\cap U^{\mathsf c}$, then $x\in F$. But this means that $$x\in\operatorname{cl}(V\cap U^{\mathsf c})=\operatorname{cl}(V\setminus U)\tag{1}.$$ Also, $$x\in\partial V= \operatorname{cl} V\cap\operatorname{cl} V^{\mathsf c}\subseteq\operatorname{cl} V^{\mathsf c}=\operatorname{cl} (V^{\mathsf c}\cup U)=\operatorname{cl} (V\setminus U)^{\mathsf c}\tag{2}.$$ Combining (1) and (2), one has that $$x\in\operatorname{cl}(V\setminus U)\cap\operatorname{cl} (V\setminus U)^{\mathsf c}=\partial(V\setminus U),$$ as desired.
Case 2: $x\notin \operatorname{cl} U^{\mathsf c}$. This case is way easier, for it can never occur. To see this, note that $U\subseteq V$ implies that $V^{\mathsf c}\subseteq U^{\mathsf c}$, which, in turn, entails that $\operatorname{cl}V^{\mathsf c}\subseteq\operatorname{cl}U^{\mathsf c}$. But $$x\in\partial V=\operatorname{cl} V\cap\operatorname{cl} V^{\mathsf c}\subseteq\operatorname{cl} V^{\mathsf c}\subseteq \operatorname{cl} U^{\mathsf c},$$ which contradicts the assumption that $x\notin \operatorname{cl} U^{\mathsf c}$. $\quad\blacksquare$
