Set of all rational points in unit ball Consider the set $A\subset \mathbb{R}^2$ of all rational points in the unit ball, that is $$A = \{(x,y)\in \mathbb{Q}^2: x^2+y^2<1\}.$$
It's easily seen that $A$ is not path-connected (since every continuous curve must pass through a point with at least one irrational coordinate, by the intermediate value theorem), so $A$ is not connected as well. Now consider the definition of connected subset:

A subset S of a topological space $X$ is connected if and only if there are no open sets $U$ and $V$ in $X$ such that $$S\subset U\cup V, S\cap U \neq \varnothing, S\cap V\neq \varnothing \text{ and } S\cap U\cap V = \varnothing.$$

Since $A$ is disconnected, there must be some open sets $U$ and $V$ in $\mathbb{R}^2$ such that $$A\subset U\cup V, A\cap U = \varnothing, A\cap V = \varnothing \text{ and } A\cap U \cap V = \varnothing.$$
Can we write specifically what $U$ and $V$ are?
Thank you very much.

There are many examples of such a $U$ and $V$.  One working example is
$$
U = \{(x,y) \in \Bbb Q^2 : x^2 + y^2 < \frac 1{\sqrt{2}} \}\\
V = \{(x,y) \in \Bbb Q^2 : \frac 1{\sqrt{2}}  < x^2 + y^2 < 1\}
$$
The only important thing about $1/\sqrt{2}$ here is that it's irrational and less than $1$.
