Connectedness of sets in the plane with rational coordinates and at least one irrational
Removing this from the unanswered list. As the comments note the OPs reasoning is correct.
To run with the comment of a proof using the definition of connectedness.
Suppose that $\mathbb{R} \backslash A$ can be written as $U \cup V$ where $U$ and $V$ are both open and disjoint. We know that there are open sets $U'$ and $V'$ in $\mathbb{R}$ such that $U = U' \cap \mathbb{R} \backslash A$ and $V = V' \cap \mathbb{R} \backslash A$. The fact that $\mathbb{R}$ is connected tells us that $U' \cap V'$ is not empty. It is clear that $U' \cap V' \subset A$ (otherwise $U \cap V$ isn't empty) so we can choose some point $x=(x_1 , x_2)$ of $A$ in $U' \cap V'$. Now $U' \cap V'$ is open in $\mathbb{R}$ so there exists some $\epsilon$ such that $B_\epsilon(x) \subset U' \cap V'$. There exists a rational $q$ such that $\vert q - x_1 \vert < \epsilon $ and then there exists an irrational number $s$ such that $s \in (q , x_1)$ and therefore $(s,x_2) \in B_\epsilon(x)$ so that $B_\epsilon(x) \cap \mathbb{R}\backslash A \not = \emptyset$ but this would mean that $U \cap V$ is not empty which is a contradiction so $\mathbb{R} \backslash A$ must be connected.