General topology Bourbaki Exercise 3 sec. 1 ch. 1 (How to use hint?) The exercise reads as follows.
Let X be a topological space. For any subset A of X, let $\alpha$(A)=$\mathring{\overline{A}}$ and $\beta$(A)=$\overline{\mathring{A}}$.
b) Show that for any subset $A$ of X, $\alpha (\alpha (A))$=$\alpha (A)$ and $\beta (\beta (A))$=$\beta(A)$.
d) Show that if A, B are two open sets such that $A \cap B=\emptyset$, then  $\alpha(A) \cap \alpha (B) = \emptyset$.  [use b)]
I solved this problem and I'm interested in the another solution of d).
This is the sketch of my solution of d)
[Let A, B are two open sets such that $A \cap B=\emptyset$. Then, for any open set $G$,
($G \subset \overline A \cap \overline B \rightarrow G \subset A \cap B=\emptyset$).
So, there's no interior points in ($\overline A \cap \overline B$) and $\alpha(A) \cap \alpha (B)= \mathring {\overline A} \cap \mathring {\overline B} = \mathring {(\overline A \cap \overline B)}=\emptyset$ ]
But I didn't use b) and I tried to think how to use b) to solve d) such as $\mathring A =A$ and  $\alpha(A) \cap \alpha (B)= \alpha (\beta (A) \cap \beta (B))$,
but I couldn't find a solution using b). How can I use b) to solve d)?
 A: This is a fairly indirect and trivial use of (b), but it’s the only one that occurs to me at the moment.
$A$ is open, so $\overline{\alpha(A)}=\beta(\overline{A})=\beta\big(\beta(A)\big)=\beta(A)=\overline{A}$. Thus, $A$ is dense in $\overline{\alpha(A)}$, and since $A\subseteq\alpha(A)\subseteq\overline{\alpha(A)}$, $A$ is dense in $\alpha(A)$. Similarly, $B$ is dense in $\alpha(B)$. Let $U=\alpha(A)\cap\alpha(B)$. $U$ is open, so $U\cap A$ is dense in $U\cap\alpha(A)=U$. Similarly, $U\cap B$ is dense in $U\cap\alpha(B)=U$. But then
$$(U\cap A)\cap(U\cap B)=U\cap(A\cap B)=\varnothing$$
is dense in $U$, so $U=\varnothing$.
A: A rather late answer to a problem which perhaps no longer presents any interest but nevertheless I wanted to point out something I consider remarkable. A very general phenomenon occurs in arbitrary topological spaces, in the following precise sense:

Proposition. Let $(X, \mathscr{T})$ be an arbitrary topological space and $U, V \in \mathscr{T}$ be two arbitrary open subsets. It then holds that $\mathring{\overline{U}} \cap \mathring{\overline{V}}=\mathring{\overline{U \cap V}}$. In a more algebraic formulation, the regularisation map given by:
\begin{align*}
\mathscr{T} &\to \mathscr{T}\\
U &\mapsto \mathring{\overline{U}}
\end{align*}
is an endomorphism of the semilattice $(\mathscr{T}, \cap)$.

Proof. It is an elementary fact that for arbitrary subsets $M, N \subseteq X$ one has $(M \cap N)^{\circ}=\mathring{M} \cap \mathring{N}$ respectively $\overline{M \cap N} \subseteq \overline{M} \cap \overline{N}$ and this entails in particular that $\mathring{\overline{U}} \cap \mathring{\overline{V}}=\left(\overline{U} \cap \overline{V}\right)^{\circ} \supseteq \mathring{\overline{U \cap V}}$, also taking into account the monotony of the interior operator. To prove our claim it thus remains to establish the converse inclusion $\mathring{\overline{U}} \cap \mathring{\overline{V}} \subseteq\mathring{\overline{U \cap V}}$.
Consider arbitrary element $x \in \mathring{\overline{U}} \cap \mathring{\overline{V}}=\left(\overline{U} \cap \overline{V}\right)^{\circ}$. By definition there exists open subset $W \in \mathscr{T}$ such that $x \in W \subseteq \overline{U} \cap \overline{V}$. In order to show that $x \in \mathring{\overline{U \cap V}}$ it suffices to establish the inclusion $W \subseteq \overline{U \cap V}$ and we proceed to do this. To this end, consider arbitrary $y \in W$. Our objective is to show that $y \in \overline{U \cap V}$ and in order to achieve it we consider arbitrary open subset $T \in \mathscr{T}$ such that $y \in T$ and attempt to show that $T \cap U \cap V \neq \varnothing$.
Since $W \subseteq \overline{U}$ we gather $y \in \overline{U}$ and since $y \in W \cap T$ and $W \cap T \in \mathscr{T}$ we have $W \cap T \in \mathscr{V}_{\mathscr{T}}(y)$ - $W \cap T$ is an open neighbourhood of $y$ with respect to $\mathscr{T}$ - and thus $W \cap T \cap U \neq \varnothing$. Therefore there exists $t \in W \cap T \cap U$. Since $t \in W \subseteq \overline{V}$ and by a token similar to the previous one $W \cap T \cap U \in \mathscr{V}_{\mathscr{T}}(t)$ is an open neighbourhood of $t$, we infer that $W 
\cap T \cap U \cap V \neq \varnothing$. Since $T \cap U \cap V \supseteq W \cap T \cap U \cap V$, it clearly follows that $T \cap U \cap V \neq \varnothing$ which means our objective is reached. This concludes the proof. $\Box$
As an immediate consequence of the above it follows that the regularisations of two disjoint open subsets remain disjoint.
