I found the following example: let $Z\subseteq\mathbb{R^2}$ the set of points with rational coordinates. The set $Z$ is disconnected, indeed a separation is given by $\{(x,y)\;|\; x<\pi\}$ and $\{(x,y)\;|\;x>\pi\}$.
Question. Why?
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Both sets are disjoint, open, and cover the entirety of $\Bbb{Q}^2$. – Theo Bendit Apr 06 '19 at 10:42
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Why are they open? – Jack J. Apr 06 '19 at 10:44
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Follow-up problems: Now consider the following sets of points --- (1) points with both coordinates irrational; (2) points with exactly one coordinate irrational; (3) points with at least one coordinate irrational; (4) points with at least one coordinate rational. – Dave L. Renfro Apr 06 '19 at 10:44
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1@JackJ The larger subsets of $\Bbb{R}^2$, ${(x, y) \in \Bbb{R}^2 : x < \pi}$ and ${(x, y) \in \Bbb{R}^2 : x > \pi }$ are open in the usual topology, as given any $(x, y)$ in the former set, the ball $B((x, y), \pi - x)$ is also contained in the former set (similar for the latter set). When you intersect these with $\Bbb{Q}^2$, this produces two relatively open subsets of $\Bbb{Q}^2$. – Theo Bendit Apr 06 '19 at 10:48