Prove that $\text{Fr}(A\cup B) = \text{Fr}(A) \cup \text{Fr}(B)$ if $ \overline{A}\cap\overline{B}=\emptyset$. I need to prove that $$\text{Fr}(A\cup B) = \text{Fr}(A) \cup \text{Fr}(B)$$
if $\overline{A}\cap\overline{B}=\emptyset$.
I have been able to prove the left to right  inclusion ($\subseteq$).
But what about the the other inclusion ($\supseteq$)?
my attempt :
$$\text{Fr}(A\cup B) \cup \text{Fr}(A\cap B)= \text{Fr}(A) \cup \text{Fr}(B).$$
So if $ \overline{A}\cap\overline{B}=\emptyset$ implies that $A\cap B= \emptyset$ and since $\text{Fr}(\emptyset)=\emptyset$ we conclude.
Is this correct?
 A: You proof is correct if you assume true (without proof)) your formula. I refer to a similar approach in this nice answer. Note that with your last approach you are proving equality immediately: you would not even need the first inclusion.
Here is another (slightly more general approach):
For the last implication, it is not even necessary to ask $\overline{A} \cap \overline{B} = \emptyset$. If $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, the result holds.
You already proved that $\partial ( A \cup B ) \subseteq \partial A \cup \partial B$.
To prove the converse (by contradiction) assume that $x \notin \partial ( A \cup B )$.  Then you can distinguish two possibilities:

*

*Either $x \notin \overline{ A \cup B } = \overline{A} \cup \overline{B}$.  In this case it comes easily that $x \notin \partial A \cup \partial B$.

*Otherwise, $x \notin \overline{ X \setminus ( A \cup B ) }$.  Then by definition $x \in X \setminus \overline{ X \setminus ( A \cup B ) } = \mathrm{Int} ( A \cup B )$.  Without loss of generality assume that $x \in A$, and since $A\cap \overline B=\emptyset$, then $x\in X\setminus\overline B$, which implies that $x\notin \partial (B)$. Furthermore, it can be shown that $U = \mathrm{Int} ( A \cup B ) \setminus \overline{B}$ is a neighbourhood of $x$ that is contained in $A$, and so $x \notin \partial A$.  So you conclude that $x \notin \partial A \cup \partial B$.

