Is $\mathbb{Q}^2$ connected? 
Is $(\mathbb Q \times \mathbb Q)$ connected?

I am assuming it isn't because $\mathbb Q$ is disconnected. There is no interval that doesn't contain infinitely many rationals and irrationals.
But how do I show $\mathbb Q^2$ isn't connected? Is there a simple counterexample I can use to show that it isn't? What would the counterexample look like?

Let $X$ be a disconnected topological space. Then $X = A \cup B$, where $A$ and $B$ are open subsets of $X$ and $A \cap B = \varnothing$. But then $X \times X = (X \times A) \cup (X \times B)$. Because $A \cap B = \varnothing$, $(X \times A) \cap (X \times B) = \varnothing$, and these sets are open by definition of the product topology.
