Let $F\subset\mathbb{Q}^2$ a closed set. Does there exists some closed and connected set $G\subset\mathbb{R}^2$ such that $F=G\cap\mathbb{Q}^2$?

For example if $F=\{a,b\}$, you can take $G$ the reunion of two lines of different irrational slopes passing through $a$ and $b$. This is a connected set and the intersection with $\mathbb{Q}^2$ is $\{a,b\}$ because the slopes are irrationnals.

But I don’t know how to prove it in general (and I don’t know if it’s true). When there are many connected components this is not clear how to connect them without adding new rational points.

Enumerate all rational points outside your set. Then cover these points by open balls by induction as follows: the next ball is centered at the first rational point not covered so far, its radius is so small that is does not intersect $F$ and the previous balls and is chosen so that the boundary of the ball does not contain rational points. Then the complement of the union of these balls is path-connected: to connect two points, draw a segment between them and go around every ball intersected by this segment.

Note that this works for any countable set, not just $\mathbb Q^2$.
