For any set $A\subseteq\mathbb{R}^n$, we have $ \overline{A^{\circ}} = \overline{\overline{A^{\circ}}^{\,\circ}}$

First, we can rephrase the question: Prove that for an open set $U\subseteq\mathbb{R}^2$, $$\overline{U}=\overline{\overline{U}^{\circ}}$$

For the direction $\subseteq$, we have $U\subseteq \overline{U}^{\circ}$ because the RHS is the maximal open set that is contained is $\overline{U}$. Since taking a closure is monotone (if $A\subseteq B$ then $\overline{A}\subseteq \overline{B}$), we get $\overline{U}\subseteq\overline{\overline{U}^{\circ}}$.

For the direction $\supseteq$, note that for any set $A$ we have $\overline{A}=\overline{\overline{A}}$. Since $\overline{U}\supseteq\overline{U}^{\circ}$ we get $\overline{U}=\overline{\overline{U}}\supseteq\overline{\overline{U}^{\circ}}$.

Interestingly, you can solve this problem by appealing to only a few abstract properties of interior and closure. See this article for more information. Here are three conditions satisfied by the closure operation:

1. $\mathrm{cl}(S) \supset S$ for all $S \subset \mathbb{R}^n$ (sets get bigger when you close them)
2. $\mathrm{cl} (\mathrm{cl} (S)) = \mathrm{cl} (S)$ for all $S \subset \mathbb{R}^n$ (the closure of a closed set is itself)
3. If $S \subset T \subset \mathbb{R}^n$, then $\mathrm{cl}( S) \subset \mathrm{cl}( T)$ (closure preserves containment)

The situation for interiors is nearly the same:

1. $\mathrm{int}(S) \subset S$ for all $S \subset \mathbb{R}^n$ (sets get smaller when you take their interiors)
2. $\mathrm{int} (\mathrm{int} (S)) = \mathrm{int} (S)$ for all $S \subset \mathbb{R}^n$ (the interior of an open set is itself)
3. If $S \subset T \subset \mathbb{R}^n$, then $\mathrm{int}( S) \subset \mathrm{int}( T)$ (taking interiors preserves containment)

Now, let $S \subset \mathbb{R}^n$ be given. We have $\mathrm{cl} (\mathrm{int}(S)) \supset \mathrm{int}(S)$. Now take the interior of both sides to get $\mathrm{int}(\mathrm{cl}(\mathrm{int}(S))) \supset \mathrm{int}(\mathrm{int}(S)) = \mathrm{int}(S)$. Taking the closure on both sides leads to $$ \mathrm{cl}(\mathrm{int}(\mathrm{cl}(\mathrm{int}(S)))) \supset \mathrm{cl} (\mathrm{int}(S))$$ which is one half what we set out to prove. I encourage you to try to deduce the reverse inclusion using similar methods beginning from $\mathrm{int}(\mathrm{cl}(\mathrm{int}(S))) \subset \mathrm{cl}(\mathrm{int}(S))$.

I like to take $ U = \operatorname{cl}(\operatorname{int}(A))$. Then $U$ is a closed set in $\mathbb{R}^n$.

So we have to prove $ U = \operatorname{cl}(\operatorname{int}(U))$.

$\operatorname{int}(U) \subset U $

As $U$ is closed and closer is smallest closed set containing $\operatorname{int}(U)$, we say $\operatorname{cl}(\operatorname{int}(U)) \subset U$.

Take any point $a \in U$.

If $a \in \operatorname{int} (U)$, trivially $a \in \operatorname{cl}(\operatorname{int} (U))$.

If $a \in \operatorname{bd}(U)$, $a$ shall be a limit point of $\operatorname{cl}(\operatorname{int}(U))$.

In any case $a \in \operatorname{cl}(\operatorname{int}(U))$.

Thus $U = \operatorname{cl}(\operatorname{int}(A))$.

Now replace $U$ by $\operatorname{cl}(\operatorname{int}(A))$, and get the result.

Replace notations only.