Are the irrationals as a subspace in the real line and in the plane a connected space? 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?

I think the confusion comes from that $\mathbb R^2 \setminus \mathbb Q^2$ is not the same thing than $(\mathbb R \setminus \mathbb Q)^2$.
In the first definition these are points of the plane which do not have both coordinates rationnal, thus there can be mixed coordinates, while in the second definition both coordinates are irrationnals.
$\mathbb R^2 \setminus \mathbb Q^2=(\mathbb R \setminus \mathbb Q)^2\quad\cup\quad(\mathbb R \setminus \mathbb Q)\times\mathbb Q\quad\cup\quad\mathbb Q\times(\mathbb R \setminus \mathbb Q)$.
But we have 
$\begin{cases}
(\mathbb R \setminus \mathbb Q)^2\cup(\mathbb R \setminus \mathbb Q)\times\mathbb Q=(\mathbb R \setminus \mathbb Q)\times\mathbb R\quad\mathrm{a\ continuous\ vertical\ path}\\
(\mathbb R \setminus \mathbb Q)^2\cup\mathbb Q\times(\mathbb R \setminus \mathbb Q)=\mathbb R\times(\mathbb R \setminus \mathbb Q)\quad\mathrm{a\ continuous\ horizontal\ path}
\end{cases}$
And this is precisely along these two sets of mixed coordinates that we can build continuous paths to connect points of $\mathbb R^2 \setminus \mathbb Q^2$.
For $(\mathbb R \setminus \mathbb Q)^2$ your construction of a positive semi-plane and a negative semi-plane works and this set is effectively disconnected.
