$A = \{ (x,0) | x \in \mathbb{Q} \} $
Now I can tell that the interior is empty and I think that the closure will be the set $A$ but with $\mathbb{R}$ instead of $\mathbb{Q}$ as $\mathbb{Q}$ is dense in $\mathbb{R}$ but I don't know how to write this as a mathematical proof and could appreciate someone showing me how.
For showing that its interior is empty it is enough to prove that $A\subseteq\mathbb R^2$ does not contain any open ball $\{(x,y)\mid d((x,y),(x_0,y_0))<r\}$ where $r>0$.
For showing that $\overline A=\{(x,0)\mid x\in\mathbb R\}$ it is enough to prove that for any $x\in\mathbb R$ no open ball centered at $(x,0)$ exists that has empty intersection with $A$, and secondly that this is not true for point $(x,y)$ if $y\neq0$.
The first proves $\supseteq$ and the second $\subseteq$.