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พอดีอยากรู้ว่าที่ผมทำถูกไหมแล้วมีวิธีอื่น อีกไหมครับ

Want to prove that $\mathbb{R}^2 $\ $\mathbb{Q}^2 $ is is connected without using path connected and countable property

Assume for contradiction that U and V form partition on $\mathbb{R}^2 $\ $\mathbb{Q}^2 $ so $U\cup V=\mathbb{R}^2 \ $\$ \mathbb{Q}^2 $ and $U\cap V=\varnothing $ and they are both open non empty subset

Let O={i}x$ \mathbb{R}$ for i $\in \mathbb{R} \ $\$ \mathbb{Q} $

Then clearly O is a subset of $\mathbb{R}^2 $\ $\mathbb{Q}^2 $ and it is connected since {i} is connected and $\mathbb{R}$ is connected so is their product. Then $(U\cap O)\cup (V\cap O) = O$ and $(U\cap O)\cap(V\cap O)=\oslash $ since U and V are disjoint by assumption .So $(U\cap O)$ and $(V\cap O)$ form a partition on O which contradicts that O is connected therefore $(U\cap V )$ is not empty imply they do not form a partition hence $\mathbb{R}^2 $\ $\mathbb{Q}^2 $ is open

มีวิธีอื่นที่ไม่ใช่ fact ว่าO is connected ไหมครับ