$A, B$ be two subsets of a metric space such that $\bar A \cap \bar B = \emptyset$, then prove that $\partial(A\cup B)=\partial A \cup \partial B$

Question

To prove the above-mentioned statement, I am just able to show $\partial(A\cup B)\subseteq \partial A \cup \partial B$ in the following manner
$\partial A \cup \partial B=$
$(\bar A \backslash A^\circ)\cup (\bar B \backslash B^\circ)=(\bar A \cup \bar B)\backslash (A^\circ \cup B^\circ)=(\overline {A \cup B})\backslash (A^\circ \cup B^\circ)\supseteq (\overline {A \cup B})\backslash (A \cup B)^\circ= \partial (A\cup B)$
$\implies \partial (A \cup B) \subseteq \partial A \cup \partial B \tag {1}\label{eq1}$
But it is general implication, I've not used the fact $\bar A \cap \bar B = \emptyset$ to prove $\eqref{eq1}$.
How to prove that $\partial A \cup \partial B \subseteq \partial(A\cup B) $?
Can anybody help me find the wayout? Thank for the assisatance in advance.
N.B. Here $\bar A = A \cup A^d$, where $A^d$ denotes the set of all limit points of $A$; $A^\circ$ denotes the set of all interior points of $A$; $\partial A$ denotes the set of all boundary points of A i.e. set of those points which are not interior nor exterior points of $A$.

Answer 1

I've not used the fact $\bar A \cap \bar B = \emptyset$ to prove $\eqref{eq1}$. How to prove that $\partial A \cup \partial B \subseteq \partial(A\cup B)$?

Well, you're kindof answering your own question there. The fact $\bar A \cap \bar B = \emptyset$ is exactly what makes the $\supseteq$ in $\overline{(A\cup B)}\setminus(A^\circ \cup B^\circ)\supseteq (\overline {A \cup B})\setminus(A \cup B)^\circ$ into $=$ because it implies $$A^\circ \cup B^\circ=(A \cup B)^\circ$$