By irrationals, $\color{blue}{\mathbb{I}}$, I mean the set $\color{blue}{\mathbb{R}\setminus \mathbb{Q}}$ and the set $\color{blue}{\mathbb{R^2}\setminus\mathbb{Q^2}}$.
My thought is no in both cases.
For the set $\color{blue}{\mathbb{R}\setminus \mathbb{Q}}$, the set $\color{blue}{\mathbb{I}}$ is the disjoint union of the negative and positive open rays starting at $0$ each intersecting the $\color{blue}{\mathbb{I}}$ (to get the two open sets in the subspace topology to form a separation).
A similar argument for the set $\color{blue}{\mathbb{R^2}\setminus \mathbb{Q^2}}$ by separating it by two open half planes along the $y$-axis.
Is this argument correct?
You're right for $\mathbb{R}\setminus\mathbb{Q}$. However, your construction doesn't work for $\mathbb{R}^2\setminus\mathbb{Q}^2$, because there are points on the $y$-axis that are in $\mathbb{R}^2\setminus\mathbb{Q}^2$ (namely, points of the form $(0,y)$ where $y$ is irrational). These points won't be in either of your open half-planes.
In fact, $\mathbb{R}^2\setminus\mathbb{Q}^2$ is actually connected; see this question.