Prove that $\text{Fr}(A\cup B) = \text{Fr}(A) \cup \text{Fr}(B)$ if $ \overline{A}\cap\overline{B}=\emptyset$.

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:

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