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?

What topology is $\mathbb{Q}^2$ equipped with? I assume you are considering the subspace topology inherited from $\mathbb{R}^2$.
To prove $\mathbb{Q}^2$ is disconnected, we can find  two disjoint open subset which constitute the whole space $\mathbb{Q}^2$. Consider the two open subset of $\mathbb{Q}^2$: $\{(x,y) \in \mathbb{Q}^2\: x < \pi\}$ and $\{(x,y) \in \mathbb{Q}^2\: x > \pi\}$. (Recall the definition of a subspace topology). They form a disjoint union of $\mathbb{Q}^2$ and are both open.
So $\mathbb{Q}^2$ is disconnected by definition.
