Use the following proposition:
If and are connected sets which are not separated, then is connected.
Choose an open disc and the closure of the other one.
Thanks all. I have another solution using this theorem: If is a connected subset of and , then is connected.
We have . Since is a connected subset of and it follows that is also connected. Thus and are connected subsets of such that . Then it follows that is connected.
That is right, you have used:
Let be a class of connected subsets of (toplogical space) with non-empty intersection. Then is connected.
And this proposition is as a consequence that any two members of are not separated. All depends on the starting point.