In my Advanced Calculus class we had this excercise and I don't know if this proof is correct.
Using the Heine-Borel Theorem, any set A in $\mathbb{R}^n$ is compact iff A it's closed and bounded. So if A is compact, I have to prove $\overline{A^°}$ is closed and bounded.
$\overline{A^°}$ is closed because it's the same as $\overline{\overline{A^°}}$. The reason behind that for any set B, $\overline{B}$ = $\overline{\overline{B}}$
Then, the only thing we have to prove is that A is bounded. If A is compact $\rightarrow$ A is bounded $\rightarrow$ $A^o$ is bounded. Now we prove that $\overline{A^°}$ is bounded. So given any $x \in \overline{A^°}$, $\forall r>0, B_r(x)\cap K^° \neq \emptyset$. So if $K^°$ is bounded, there exists $\epsilon$ such as $K^° \subseteq B_\epsilon (x)$. But if $x \in \overline{A^°}$, there is $\tilde\epsilon > \epsilon > 0$ such as $ B_\epsilon(x)\subseteq B_\tilde\epsilon(x)$. Because this is for every $x \in \overline{A^°}$, we can bound every element of $ \overline{A^°}$ and the union of all the balls is an open set that bounds $x \in \overline{A^°}$.
Is this right?
It is easier than you are making it. The closure of a set is the intersection of all closed sets containing the set. Since the interior of a compact set is a subset of the compact set and the compact set is closed, the intersection of all closed sets containing the interior, i.e. the closure, is a subset of the compact set. Since the compact set is bounded, so is the closure. Therefore, the closure of the interior is compact.