TITLE: If $A$ and $B$ are two disjoint non empty subsets of $\mathbb{R}^2$ such that $A\cup B$ is open....
QUESTION [8 upvotes]: Question is  :
If $A$ and $B$ are two disjoint non empty subsets of $\mathbb{R}^2$ such that $A\cup B$ is open in $\mathbb{R}^2$ . Then which of the following is true?

If $A$ is open and $A\cup B$ is connected then $B$ must be closed in $\mathbb{R}^2$.
If $A$ is closed then $B$ must be open in $\mathbb{R}^2$.
If both $A$ and $B$ are connected, then $A\cup B$ must be disconnected.
If  $A\cup B$ is disconnected then both $A$ and $B$ are open.

I am sure that first and second bullet are correct but do not know how to make it more clear.
For first bullet, 
Suppose $B$ is open then there is no chance of $A\cup B$ being connected though $A$ and $B$ are connected. (As $A$ and $B$ are disjoint) But $B$ being Not open does not imply $B$ being closed. So, I would not say this is a proof but i would see this can be made to a proof.(I guess it can)
For second bullet,
Suppose $B$ is also closed then $A\cup B$ is closed but we have given that $A\cup B$ is open. But $B$ being Not closed does not imply $B$ being open. So, I would not say this is a proof but i would see this can be made to a proof.(I guess it can)
For third bullet,
I can surely say that this is false. Suppose $A\cup B$ is connected then $A$ and $B$ should have a common point But $A$ and $B$ are disconnected.Thus third option is false.
For fourth bullet,
I think it is true.In general, $A\cup B$ being disconnected does not imply  both $A$ and $B$ are open. but we have a condition $A\cup B$ is open. I feel this would force both $A$ and $B $ to be open. I could not make this more clear.
I would be thankful if someone can help me to clear this.
Thank you :)

REPLY [6 votes]: The problem with your approach is that a set can be neither open nor closed. So you cannot prove the contrapositive by turning open into closed.
For the first bullet, consider $A = (0, 1) \times (0, 1)$, $B = [1, 2) \times (0, 1)$. Then $A \cup B = (0, 2) \times (0, 1)$.
For the second bullet, consider $(A \cup B) \cap A^\complement$. What is it equal to? Why is it open?
For the third and fourth bullets, use examples similar to the first bullet.