Prove that $\partial (A \cup B)= \partial(A) \cup \partial(B)$ if $\bar {A} \cap \bar {B} = \phi$. The question is $:$
Let $(X,d)$ be a metric space.Let $A,B \subseteq X$.Then prove that $\partial (A \cup B)= \partial(A) \cup \partial(B)$ if $\bar {A} \cap \bar {B} = \phi$ , where $\partial (A)$ and $\bar A$ respectively denote the boundary of $A$ and closure of $A$ in $(X,d)$.
My attempt $:$
Let $A^{o}$ denote the interior of $A$.Then we know that $\partial(A) = \bar A \setminus A^{o}$. Now,$\partial (A \cup B) = \overline {A \cup B} \setminus (A \cup B)^{o} = (\bar A \cup \bar B) \setminus (A \cup B)^{o} = (\bar A \setminus (A \cup B)^{o}) \cup (\bar B \setminus (A \cup B)^{o}) \subseteq (\bar A \setminus A^{o}) \cup (\bar B \setminus B^{o}) = \partial(A) \cup \partial(B)$.
But I fail to prove the other part of the inequality using the given condition.
Please help me in proving it.
Thank you in advance. 
 A: Suppose $x\in \partial (A)\cup\partial(B)$. Assume WLOG $x\in \partial (A)$ (because the two boundaries are disjoint).  Then $\forall \epsilon>0, B(x,\epsilon)\cap A\neq \emptyset\ \text{and}\ B(x,\epsilon)\cap(X\setminus A)\neq\emptyset$. Therefore $\forall \epsilon>0, B(x,\epsilon)\cap (A\cup B)\neq\emptyset$. Assume $\exists\epsilon_0>0$ such that $B(x,\epsilon_0)\cap (X\setminus(A\cup B))=\emptyset$. Then $B(x,\epsilon_0)\subseteq A\cup B$. Therefore $x\in\ \text{int} (A\cup B)$. Therefore $x \in\text{int}B$. So $\exists \epsilon_1>0$ such that $B(x,\epsilon_1)\subseteq B$. Therefore $B(x,\epsilon_1)\cap A =\emptyset$; contradiction. So we have $\forall\epsilon>0,B(x,\epsilon_0)\cap (X\setminus(A\cup B))\neq\emptyset$. Hence $x\in\partial(A\cup B)$.
A: @ArnabChatterjee Your logic looks sound.  Excellent reasoning!  
A ${slightly}$ more direct approach you might consider could be as follows:
Let $ x \in \partial{A}\cup\partial{B}.$  WLOG $ x \in \partial{A}$ (just as before).  Since $cl(A) $ and $cl(B) $ are disjoint, then $cl(A) \subseteq  X \backslash cl(B)$, which is open.  Then there must be an $\epsilon_0$ that keeps $B(x,\epsilon_0) $ disjoint from $cl(B)$.  Since $x$ is in the boundary of $A$, then $B(x,\epsilon_0)$ has non-empty intersection with $ A \subseteq A \cup B$, as well as non-empty intersection with $X \backslash A$.  It follows from such $\epsilon_0$ that any $y$ in $B(x,\epsilon_0)\backslash A$ cannot be in $B$, so such $y$ is also in $X\backslash B $; that is to say that $B(x,\epsilon_0) \cap \color{red}{(X\backslash A)\cap (X\backslash B)}$$ \:(which\:is\: \color{red}{X\backslash(A\cup B)})$ is also non-empty.  Hence, for any $\epsilon \leq \epsilon_0 $ (i.e. $\forall \epsilon > 0$), we have $B(x,\epsilon)\cap(A\cup B)\neq \emptyset$, and $B(x,\epsilon)\cap [X\backslash(A\cup B)]\neq \emptyset$.
