Let $A \subset \mathbb{R}^n$.
Prove that if $A \subset \bigcup\limits_{\alpha \in J} U_{\alpha}$, while $U_{\alpha}$ are open sets,
then it is possible to have a countable union $\{U_{\alpha_{n}}\}_{n\in \mathbb{N}}$, such that: $A \subset \bigcup\limits_{n\in \mathbb{N}} U_{\alpha_{n}}$.
Please any help or direction.
Fix a countable basis $\{ B_k \mid k \in \mathbb N\}$ of $\mathbb R^n$. Fix $X \subseteq \mathbb N$ such that $$ \bigcup_{\alpha \in J} U_{\alpha} = \bigcup_{k \in X} B_k $$ and such that for every $k \in X$ there is some $\alpha \in J$ with $B_k \subseteq U_\alpha$. (This is possible because $\bigcup_{\alpha \in J} U_{\alpha}$ is open and $\{B_k \mid k \in \mathbb N\}$ is a basis.)
$X$, combined with a fixed well-order on $J$, induces a natural countable subset of $J$ as desired -- I'll leave it to you to figure out how.