Question: Here's an interesting problem, and result, that I wish to share with the math community here at Math SE. I think I've found a proof without words...

[![enter image description here](https://i.stack.imgur.com/rcIcf.jpg)](https://i.stack.imgur.com/rcIcf.jpg)

I came across this problem sometime back, and here's a possible proof without words that I'll post as a picture - 

[![enter image description here](https://i.stack.imgur.com/KSbLS.jpg)](https://i.stack.imgur.com/KSbLS.jpg)

*I claim* that one can, for any inscribed quadrilateral EFGH, make necessary constructions (several parallelograms) as in the figure, and hence show that at least one of the diagonals is parallel to a side of the quadrilateral ABCD. 

Is there anything I'm missing, or is the proof complete? Let me know! 

Also, please post other proofs in the answers section! (So we can all solve the problem together and discuss several methods for the same - it'll be of use to everyone to know all possible methods of approaching this problem) 

I was thinking about a possible solution using complex numbers, assuming one of the side pair to be parallel to the real axis in Argand's plane, for simplicity. A similar solution, not involving too much calculation can be done using vectors! 
I'm not sure about coordinate geometry, as it tends to get quite cumbersome in such situations, however if anyone does manage to prove it neatly, please post the solution.

**Share your ideas!**

Answer: 
**This is a hint, not a solution.**

Hint: Let's call your set $X$, just to give it a name. 

1. Can you show that $\Bbb Q$ is disconnected by exhibiting nonempty open sets $U$ and $V$ with $\Bbb Q = U \cup V$ and $U \cap V = \emptyset$?
2. If you can, then $U\_1 = U \times \Bbb R$ and $V\_1 = V \times \Bbb R$ are both open. (Why?)
3. $U' = U\_1 \cap X$, and $V' = V\_1 \cap X$ are both open subsets of $X$ in the subspace topology. (Why?)
4. $U'$ and $V'$ are nonempty.

You take it from here.
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.