Let $\{x_n\}_{n=1}^{\infty}$ be an enumeration of $\Bbb Q$, $\{y_\lambda\}_{\lambda \in \Lambda}$ an enumeration of $\Bbb R\setminus\Bbb Q$.
Proof: for any families $\{r_n\}_{n=1}^\infty,\,\{ \rho_\lambda\}_{\lambda\in\Lambda}$ of positive reals, we have: $$\left ( \bigcup_{n=1}^{\infty} B((x_n,r_n),r_n) \right ) \bigcap \left ( \bigcup_{\lambda \in \Lambda}B((y_{\lambda},\rho_{\lambda}),\rho_{\lambda}) \right )\ne \emptyset $$
Where $B$ denotes open balls in $\Bbb R^2$.
And my teacher gives a hint: try to use Baire category theorem.
I guess that it is necessary to use the method of proof by contradiction, and then construct a "complete metric space" so that it is not a Baire space to obtain the contradiction. But I have no idea how to construct one.
My teacher gave the answer recent day:
Assume that we got both $\{r_n\}_{n=1}^{\infty}$ and $\{\rho_{\lambda}\}_{\lambda \in \Lambda}$ such that $$(\bigcup_{n=1}^{\infty}S_n) \bigcap (\bigcup_{\lambda \in \Lambda}T_{\lambda} )=\emptyset$$ hold.
where $S_{i}=B((x_n,r_n),r_n)$ and $T_{\lambda}=B((y_{\lambda},\rho_{\lambda}),\rho_{\lambda})$ .
Set $E_k=\{ y_{\lambda} \in \Bbb R\setminus\Bbb Q:\rho_{\lambda}\ge \frac{1}{k}\}$,then we have $$\Bbb R\setminus\Bbb Q=\bigcup_{k=1}^{\infty}E_{k}$$ and$$\Bbb R=\Bbb Q \cup(\Bbb R\setminus\Bbb Q)=\left (\bigcup_{n=1}^{\infty}x_n \right )\bigcup \left (\bigcup_{n=1}^{\infty}E_k \right )$$ hold.
By Baire category theorem we know that there exists $k_0\in \Bbb N^{+}$ such that $E_{k_0}$ is a nowhere dense set in $\Bbb R$,which mean there exist a open set $(\Bbb R \ni)A \subset \overline{E_{k_0}}$ i.e. $E_{k_0}$ is dense in $A$.
Due to $A \in \Bbb R$ ,The structure theorem for open sets and the truth that $\Bbb Q$ is dense in $\Bbb R$ we are sure that $A \cap \Bbb Q \ne \emptyset$.
Then accroding to the discussing above , for $x_m \in A\cap \Bbb Q \subset \Bbb Q$ and $\forall \epsilon<2\sqrt{\frac{r_m}{k_0}}$ , there exists $y_{\lambda_m}\in E_{k_0}\subset \Bbb R \setminus\Bbb Q$ such that $|y_{\lambda_m}-x_m|<\epsilon$, so we got $$(y_{\lambda_m}-x_m)^2+(\rho_{\lambda_m}-r_m)^2<(\rho_{\lambda_m}-r_m)^2+\epsilon^2<(\rho_{\lambda_m}-r_m)^2+\frac{4r_m}{k_0}<(\rho_{\lambda_m}-r_m)^2+4r_m \rho_{\lambda_m}=(\rho_{\lambda_m}+r_m)^2$$ which means that $S_m \cap T_{\lambda_m} \ne \emptyset$,contradicts.