Let $X$ be a topological space and $A \subset X$ a constructable subset. Therefore there exist open $O_i$ and closed $A_j$ such that ${\displaystyle A=\bigcup _{i=1}^{n}(O_{i}\cap A_{i})}$.
I'm looking for a proof of following statement:
$A$ is open (in $X$) iff $A$ is closed under generalization.
Remark: The latter means that if $a \in A$ and $b \in X$ with $a \in \overline{b}$ then $b \in A$.
On
This is not true without further assumptions - notably, that $X$ is sober and noetherian. In the case when these assumptions are made, this is Stacksproject tag 0542 which I have previously commented on a question of yours.
Counterexample in the case that $X$ is not noetherian: consider $\Bbb R$ with the usual topology. Then every set is stable under generalization because there are no nontrivial generalizations nor specializations. On the other hand, $[0,1)=(-1,1)\cap [0,2]$ is constructible but not open.
Counterexample in the case that $X$ is not sober: Let $R=\overline{\Bbb F_2}$ and let $X=R^2$ with the closed sets given by the Zariski topology. It is well known that this is noetherian and not sober (there are no generic points, so it is not true that every irreducible closed subset is the closure of a unique point), and there are no nontrivial specializations or generalizations. Pick any nontrivial intersection of an open and closed set and you have a counterexample.
I'm not sure that's true for general topological spaces. However, it is true for "Noetherian sober" spaces, which should be enough for the application you have in mind? (I saw your previous question about the Going-Down Theorem!)
Anyway, the direction that's omitted in the stacks project is easy: Suppose $A$ is open. Let $a \in A, \ b \in X$. If $b \notin A$, then $b \in X \setminus A$ (which is closed), hence $\overline{\{b\}} \subseteq X \setminus A$ and $a \notin \overline{\{b\}}$. Therefore, if $a \in \overline{\{b\}}$, then $b \in A$.
The converse is proven in Stacks (Lemma 5.19.10): https://stacks.math.columbia.edu/tag/0060