Let $A\cup B$ be open, disconnected in $\Bbb{R}^2$ where $A,B$ are non-empty, disjoint. Are both $A,B$ open in $\Bbb{R}^2$?

Question

I have tried it in the following manner-
Assume $A$ is not open. Then $\exists x\in A$ such that $x\notin A^\circ$ i.e. $\forall \epsilon>0, B(x,\epsilon)\not\subset A$ .
Now $x\in A\cup B$, open in $\Bbb{R}^2$. Hence $\exists r>0$ such that $B(x,r)\subset A\cup B$. But $B(x,r)\not\subset A$, so we must have $B(x,r)\cap B\neq\emptyset$. Again for any $0<\epsilon\le r, B(x,\epsilon)\subset A\cup B$ and $B(x,\epsilon)\not\subset A$, so we must have $B(x,\epsilon)\cap B\neq \emptyset$ for all $0<\epsilon\le r$. Hence $x\in \overline{B}$.
$\therefore x\in A\cap\overline{B}\implies A\cap \overline{B}\neq \emptyset$.
Now if both $A$ and $B$ are connected then $A\cup B$ will be connected (but here it is given that $A\cup B $ is disconnected). So we must have either $A$ or $B$ is disconnected.
Now from this stage I cannot proceed further to get a contradiction. Can anyone guide me to conclude? Thanks for help in advance.

Answer 1

It's not true. For example, $A = ((0,1] \cup (3,4)) \times \mathbb R$, $B = ((1,2) \cup [4,5)) \times \mathbb R$, so $A \cup B = ((0,2) \cup (3,5)) \times \mathbb R$ is open and disconnected, but neither $A$ nor $B$ is open.

Answer 2

Let

$$A=\big((-1,0)\times(0,1)\big)\cup\left\{\left\langle 0,\frac12\right\rangle\right\}\cup\big((-1,0)\times(-1,0)\big)$$

and

$$B=\left(\big([0,1)\times(0,1)\big)\setminus\left\{\left\langle 0,\frac12\right\rangle\right\}\right)\cup\big((0,1)\times(-1,0)\big)\;.$$

Then $A\cap B=\varnothing$,

$$A\cup B=\big((-1,1)\times(0,1)\big)\cup\big((-1,0)\times(-1,0)\big)\cup\big((0,1)\times(-1,0)\big)$$

is open and disconnected in $\Bbb R^2$, and neither $A$ nor $B$ is open.