Is below subspace is connected? Question: the subspace $\mathbb{Q}×[0,1]$ of $\mathbb{R^2}$(with usual topology) is connected? 
I know the definition of connected set but yet I had solve the examples of $\mathbb{R}$. I know $\mathbb{Q}$ is totally disconnected. So I think given subspace will be disconnected. But, this is not the proof! To prove: the subspace $\mathbb{Q}×[0,1]$ is disconnected, we have to express it as disjoint union of two non-empty open sets in subspace topology! But as a beginner, I can't able do this!
please anyone help me to prove it is disconnected! I am stuck on this from hours and i don't want to skip proof.
 A: This is a hint, not a solution.
Hint: Let's call your set $X$, just to give it a name. 


*

*Can you show that $\Bbb Q$ is disconnected by exhibiting nonempty open sets $U$ and $V$ with $\Bbb Q = U \cup V$ and $U \cap V = \emptyset$?

*If you can, then $U_1 = U \times \Bbb R$ and $V_1 = V \times \Bbb R$ are both open. (Why?)

*$U' = U_1 \cap X$, and $V' = V_1 \cap X$ are both open subsets of $X$ in the subspace topology. (Why?) 

*$U'$ and $V'$ are nonempty. 
You take it from here. 
A: Looking over the comments to John Hughes' answer, I see that you were advised to study how the definitions and properties of connected\subspace\product play out together.
With that as the background, the problem can be tackled by finding a nonempty and proper subset of $D = \mathbb{Q}×[0,1]$ that is both open and closed.
So we want to find an open set $U$ of $\mathbb{R^2}$ and a closed set of $C$ of $\mathbb{R^2}$ such that
$\tag 1 \emptyset \ne U \cap D = C \cap D \ne D$
There are two candidates that practically 'jump out' for inspection:
$\quad U = (\sqrt2, +\infty) \times \mathbb R$
and
$\quad C = [\sqrt2, +\infty) \times \mathbb R\;$ (recall that the product of two closed sets is closed)
We leave it to the OP to wrap up any loose ends.
