Is this set containing all circles of rational radius connected? Given $r>0$, let $C_r=\{(x,y)\in \mathbb{R}^2:x^2+y^2=r^2\}$. Let $X$ be a subset of $\mathbb{R}^2$ having the following properties:

*

*For all $r\in \mathbb{Q}$, $C_r\subset X$.

*For all $r\in \mathbb{R}-\mathbb{Q}$, $X\cap C_r\neq \emptyset$.

Is $X$ connected?
My attempt:
I tried to prove that $X$ is not connected by considering, $A=\bigcup_{r\in \mathbb{Q}} C_r,$ and $B=X-A$. But I've not been able to prove that $A$ and $B$ are separated sets, that is, $\overline{A}\cap B=A\cap \overline{B}=\emptyset$.
 A: Edit: I found a proof to see $X$ is connected. Suppose $X\subset A\cup B$, where $X\cap A\neq \emptyset$, $X\cap B\neq \emptyset$ and $A\cap B=\emptyset$, with $A,B$ open subsets of $\mathbb{R}^2$. Suppose wlog that $C_1\subset A$, and let $a=\inf \{r^*>0 \,|\, \cup_{r^*<r<1, r\in \mathbb{Q}} C_r\subset A\}$, certainly $a<1$ because $C_1$ is compact and hence has a finite open cover contained in $A$. Suppose $a>0$, if $a$ is rational then $C_a$ can not be contained in $A$ since this would yield a contradiction: there exists a Lebesgue number $l>0$ such that every ball of radius $l$ centered in any point of $C_a$ is contained in $A$, then $A$ contains an open set with form of ring  which also contains a circumference of radius less than $a$; therefore $C_a\subset B$, but since $C_a$ is compact there is a finite open cover of $C_a$ contained in $B$ which also yields a contradiction, because $C_{a+\varepsilon}\subset A$ for small $\varepsilon>0$ such that $a+\varepsilon$ is rational. Hence $a$ is irrational, but then there exists a $(x,y)\in C_a \cap X$, supposing $(x,y)\in A$ would yield a contradiction because $A$ would have to contain a circumference of rational radius and less than $a$, using circumferences are connected, on the contrary if $(x,y)\in B$ would be also a contradiction, since $B$ would have to intersect circumferences contained in $A$. It follows that $a=0$. Now let $b=\sup\{r^*>0\,|\, \cup_{1<r<r^*,r\in \mathbb{Q}} C_r \subset A\}$, similarly one gets that $b=+\infty$. Therefore $A$ contains all circumferences of rational radius. As every point of $\mathbb{R}^2$ is an acummulation point of $A$, since it contains a dense subset, no point can be contained in $B$ which is also open, hence $B=\emptyset$ and $X$ is connected.
A: Any continuous function $f:X\rightarrow\{T,F\}$ would induce a continuous function $g:\mathbb{R^+}\rightarrow\{T,F\}$ defined by $g(x)=f(C_x\cap X)$. Notice the image of $g$ equals the image of $f$.
