Please verify if what I did is correct, I'm insecure specially at the derived part.
$A = [0,1] \cap \mathbb{Q}$
$A \subseteq \mathbb{Q} \implies A^0 \subseteq \mathbb{Q}^0 = \emptyset \implies A^0 = \emptyset$ ($\mathbb{Q}^0$ is empty because any ball centered in a rational contains irrational points, by density)
$\overline{A}=\overline{[0,1]}\cap\overline{\mathbb{Q}}=[0,1]\cap\mathbb{R}=[0,1]$
$\partial A = \overline{A} \setminus A^0 = [0,1] \setminus \emptyset = [0,1]$
$A' = \overline{A} \setminus A = [0,1] \setminus ([0,1] \cap \mathbb{Q}) = [0,1] \cap (\mathbb{R} \setminus \mathbb{Q})$
But if I take a rational $r \in [0,1]$, wouldn't any ball centered at it have nonempty intersection with $[0,1]$ and in different points than $r$, because of the density again? I'm confused...
Thanks.