Prove $\partial (A \cup B)\subset \partial A\cup\partial B$ How to use this definition of boundary: "any open ball centered at a boundary point of a set $A$ intersects both $A$ and $A^C$" to prove $\partial (A \cup B)\subset \partial A\cup\partial B$?
I tried by arguing that it is equivalent to: $\partial (A \cup B)\subset \partial A$ or  $\partial (A \cup B)\subset \partial B$. And assume  $\partial (A \cup B)\not\subset \partial A$ then we need to show $\partial (A \cup B)\subset \partial B$ must hold. I don't know how to use the definition to prove this.
 A: $\forall x\in\partial(A\cup B)$ and $\forall r > 0$, we have : 
$$B(x,r)\cap(A\cup B)=(B(x,r)\cap A)\cup(B(x,r)\cap B)\ne \emptyset\,...(1)$$
$$B(x,r)\cap(A\cup B)^C=(B(x,r)\cap A^C)\cap(B(x,r)\cap B^C)\ne \emptyset\,...(2)$$
By (2), we have $B(x,r)\cap A^C\ne \emptyset$, and $B(x,r)\cap B^C\ne \emptyset$
By (1), we have $B(x,r)\cap A$ and $B(x,r)\cap B$ cannot be both $\emptyset$
Thus, we could conclude that at least one of the following two claim (*, and **) would be true:
$$B(x,r)\cap A\ne \emptyset\, and\, B(x,r) \cap A^C\ne \emptyset\,...(*)$$
$$B(x,r)\cap B\ne \emptyset\, and\, B(x,r) \cap B^C\ne \emptyset\,...(**)$$
And that is to say $x\in \partial A \cup\partial B$
Thus $\partial(A\cup B) \subset\partial A \cup\partial B$
Attach a picture below for intuition. $\partial(A\cup B)$ is only the blue part, and $\partial A \cup\partial B$ consists of both the blue and red parts.

A: Let $p\in \partial(A\cup B)$
Let $B_r(p)$ be an open ball centered at $p$ that intersects both $A\cup B$ and $(A\cup B)^c=A^c\cap B^c$
Since $B_r(p)$ intersects $A\cup B$ it intersects $A$ or it intersects $B$.
Since it intersects $A^c\cap B^c$ it intersects both $A^c$ and $B^c$.
Without loss of generality, assume $B_r(p)$ intersects $A$. Then since it intersects $A^c$, $p\in\partial A\subset\partial A\cup\partial B$. Since this is true for all $p\in\partial(A\cup B)$, $\partial(A\cup B)\subset\partial A\cup\partial B$.
A: Suppose $p \notin \partial A \cup \partial B$.
This means that $p \in A^\circ$ or $p \in (X\setminus A)^\circ$
and $p \in B^\circ$ or $p \in (X\setminus B)^\circ$.
If the first option holds for one of them, $p \in A^\circ \subset (A \cup B)^\circ$ or $p \in B^\circ \subset (A \cup B)^\circ$ and then $p \notin \partial(A \cup B)$.
If then both last options hold, $$p \in (X\setminus A)^\circ \cap (X\setminus B)^\circ = ((X \setminus A) \cap (X \setminus B))^\circ = (X\setminus (A \cup B))^\circ$$ and again $p \notin \partial(A \cup B)$.
This proves the contrapositive. We only use $(C \cap D)^\circ = C^\circ \cap D^\circ$ in all spaces and for all subsets. And I use that by definition $X = A^\circ \cup \partial A \cup (X \setminus A)^\circ$ and the union is disjoint.
A: Be defnition
$$\partial A=\overline{A}\cap\overline{X\setminus A}=\partial(X\setminus A)$$
Let's denote $A^c=X\setminus A$. Recall that for any sets $A,B,C,D$ in a topological space

*

*$\overline{A\cup B}=\overline{A}\cup\overline{B}$.

*$\overline{C\cap D}\subset\overline{C}\cap\overline{D}$
Then
$$\begin{align}
\partial(A\cup B)&=\overline{A\cup B}\cap \overline{A^c\cap B^c}\\
&\subset \Big(\overline{A}\cup\overline{B}\Big)\cap\Big(\overline{A^c}\cap\overline{B^c}\Big)\\
&=\Big(\overline{A}\cap\big(\overline{A^c}\cap\overline{B^c}\big)\Big)\cup\Big(\overline{B}\cap\big(\overline{A^c}\cap\overline{B^c}\big)\Big)\\
&\subset\big(\overline{A}\cap\overline{A^c}\big)\cup\big(\overline{B}\cap\overline{B^c}\big)
\end{align}$$
