Basic properties of closure and interior I have proved the following $$\overline{A\cup B}=\overline{A}\cup\overline{B}$$ $$\overline{A\cap B}\subset\overline{A}\cap\overline{B}$$
Now, to prove $$\mathring{A\cup B}\supset\mathring{A}\cup\mathring{B}$$ $$\mathring{A\cap B}=\mathring{A}\cap\mathring{B}$$ I want to use the fact that $(\overline{A})^{c}=\mathring{(A^{c})}$ and $(\mathring{A)}^{c}=\overline{A^{c}}$. 
For instance, 
\begin{align*}
\overline{A\cup B}=\overline{A}\cup\overline{B}\iff & (\overline{A\cup B})^{c}=(\overline{A}\cup\overline{B})^{c}\\
\iff & \mathring{\widehat{(A\cup B)^{c}}}=\overline{A}^{c}\cap\overline{B}^{c}\\
\iff & \mathring{\widehat{A^{c}\cap B^{c}}}=\mathring{\widehat{A^{c}}}\cap\mathring{\widehat{B^{c}}}\\
\iff & \mathring{A\cap B}=\mathring{A}\cap\mathring{B}
\end{align*}
My question is: what guarantees me the validity of the last $\iff$?
 A: The "for all". You have proved that for all $A,B$ you have $\overline{A\cup B} = \overline{A} \cup \overline{B}$.
Given $C,D$, apply it to $A = C^c$ and $B = D^c$, and taking complements yields $(C\cap D)^{\Large\circ} = \overset{\Large\circ}{C} \cap \overset{\Large\circ}{D}$. Now rename $C$ to $A$ and $D$ to $B$.
A: $$[\text{int}(A\cap B)]^{\complement}=\overline{(A\cap B)^{\complement}}=\overline{A^{\complement}\cup B^{\complement}}=\overline{A^{\complement}}\cup\overline{B^{\complement}}=[\text{int}(A)]^{\complement}\cup[\text{int}(B)]^{\complement}$$
where the third equality is the one you allready proved.
Then:$$\text{int}(A\cap B)=[[\text{int}(A)]^{\complement}\cup[\text{int}(B)]^{\complement}]^{\complement}=\text{int}(A)\cap\text{int}(B)$$
A: I'd write it differently, using $\operatorname{int}(A)$ insteas of $A^\circ$ for clarity, and $$X\setminus\overline{A} = \operatorname{int}(X\setminus A)$$
which is equivalent to $$\overline{A} = X\setminus (\operatorname{int}(X\setminus A))$$
So $$\overline{A \cup B} = X\setminus \left(\operatorname{int}(X\setminus (A \cup B)\right) =
X\setminus (\operatorname{int}((X\setminus A) \cap (X\setminus B)))=
X \setminus (\operatorname{int}(X\setminus A) \cap \operatorname{int}(X\setminus B)) = (X\setminus \operatorname{int}(X \setminus A) )\cup (X\setminus \operatorname{int}(X \setminus B) )  =\overline{A} \cup \overline{B}$$
applying de Morgan twice and the intersection equality for interior once.
The inclusion goes in the same way; recall that $A \subseteq B$ iff $X\setminus A \supseteq X\setminus B$ etc.
THis shows how to derive the closure properties from the interior properties; the other way round is the same using 
$$\operatorname{int}(A) = X\setminus (\overline{X\setminus A})$$
etc. Just direct computations using de Morgan twice, again.
