In Example 6.17 in Section 6.3 of Linear Integral Equations by Rainer Kress, Kress uses that
$$\partial ((B_r(x))^C\cap D)=(\partial B_r(x)\cap D)\cup(\partial D\cup B_r(x))\qquad \text{(EQ 1)}$$ for an arbitrary bounded domain $D\in \mathbb{R}^n$ and $x\in \partial D$ and without proving this statement.
Drawing a picture when $D$ is 2-dimensional the result seems obvious but I want to prove the result rigorously.
I started the proof by rewriting the left-hand side of (EQ 1) and got
$$\partial(B_r(x)^C\cap D)=\overline{B_r(x)^C\cap D}\setminus(B_r(x)^C\cap D)^0$$
$$=(\overline{B_r(x)^C\cap D})\cap((B_r(x)^C\cap D)^0)^C$$
$$=(\overline{B_r(x)^C\cap D})\cap((B_r(x)^C)^0\cap D)^C$$
$$=(\overline{B_r(x)^C\cap D})\cap(((B_r(x)^C)^0)^C\cup D^C)$$
$$=(\overline{B_r(x)^C\cap D})\cap(\overline{(B_r(x)^C)^C}\cup D^C)$$
$$=(\overline{B_r(x)^C\cap D})\cap(\overline{B_r(x)}\cup D^C)$$
$$=(\overline{B_r(x)^C\cap D})\cap(\overline{B}_r(x)\cup D^C)\qquad \text{(EQ 2)}$$
Then I rewrote the right-hand side to get
$$(\partial B_r(x)\cap D)\cup (\partial D\cap B_r(x))=((\overline{B}_r(x)\setminus B_r(x))\cap D)\cup((\overline{D}\setminus D)\cap B_r(x)^C)$$
$$=(\overline{B}_r(x)\cap B_r(x)^C\cap D)\cup (\overline{D}\cap D^C\cap B_r(x)^C)\qquad \text{(EQ 3)}$$
Show that the left-hand side of (EQ 1) is included in the right-hand side of (EQ 1), I use the property that contains the intersection of the closures of two sets contains the closure of an intersection of the two sets and EQ1-EQ 2:
$$\partial(B_r(x)^C\cap D)=(\overline{B_r(x)^C\cap D})\cap(\overline{B}_r(x)\cup D^C)$$
$$\subset (\overline{B_r(x)^C}\cap \overline{D})\cap (\overline{B}_r(x)\cup D^C)$$
$$=(B_r(x)^C\cap \overline{D})\cap(\overline{B}_r(x)\cup D^C)$$
$$=B_r(x)^C\cap \overline{D}\cap \overline{B}_r(x)\cup B_r(x)^C\cap \overline{D}\cap D^C$$
$$=(\partial B_r(x)\cap D)\cup (\partial D\cap B_r(x))$$
For the reverse inclusion, I would need
$$(\overline{B_r(x)^C\cap D})\cap(\overline{B}_r(x)\cup D^C)\supset (\overline{B_r(x)^C}\cap \overline{D})\cap (\overline{B}_r(x)\cup D^C).$$
This would imply that
$$(\overline{B_r(x)^C\cap D})\supset (\overline{B_r(x)^C}\cap \overline{D}) \qquad \text{(EQ 4)}$$
It not generally true that the closure of the intersection of the closures of two sets is contained in the closure of their intersection (for example, consider the sets $(0,1)$ and $(1,2)$).
What separates this from the most general case are that $B_r(x)^C$ is closed, $D$ is open, and $\bar{D}$ is compact (since any closed and bounded subset of $R^n$ is compact).
I'm really hoping that these two assumptions make EQ 4 true, but I don't see a way to prove this.
I didn't know what to title this question but the point of the question essentially boils down to analyzing the closure of the intersection of a closed set and an open set which has compact closure.
EDIT: What I am trying to prove is trivially false if $D=B_r(x)$. This means that I must use the assumption that $x\in \partial D$ somehow.
Best Answer
Take D to be a punctured disc, missing a point p, and B to be a ball centered on the boundary of D and with p on its boundary. p is in the LHS but not the RHS, whether you rewrite the equation as user87690 suggests, or just by changing the last $\cup$ to $\cap$, or leave it as it is.