Possibility of Union of two connected sets being connected with empty Intersection I know that :

If $A$ and $B$ are connected  then $A\cup B$ is also connected if $A\cap B\neq \emptyset$

My Question is.. 

Are there any examples where $A$ and $B$ are connected and $A\cup B$ is connected even though $A\cap B=\emptyset$ 

I have seen one possibility 
$S=\{(x,y):x>0; y=\sin (\frac{1}{x})\}\cup \{(0,y) : -1\leq y\leq 1\}$ as subsepace of $\mathbb{R}^2$
This would be mostly called as Topologist sine curve if i am not wrong...
$\{(x,y):x>0; y=\sin (\frac{1}{x})\}$ is connected 
and $\{(0,y) : -1\leq y\leq 1\}$ is connected with
$\{(x,y):x>0; y=\sin (\frac{1}{x})\}\cap \{(0,y) : -1\leq y\leq 1\}=\emptyset$
I would like to see if there is anything nice about this example and i would like to see if there are any other examples preferably non trivial.
Thank you. 
 A: A pretty trivial example would be $A = [0,1],\; B = (1,2]$. But that trivial example captures the essence already. We have the
Theorem: Let $A$ be connected, and $A \subset E \subset \overline{A}$. Then $E$ is connected.
From that we derive the
Proposition: If $A$ and $B$ are nonempty connected sets with $A\cap B = \varnothing$, then $A\cup B$ is connected if and only if $A\cap \overline{B}\neq \varnothing$ or $\overline{A}\cap B \neq \varnothing$.
For if e.g. $A \cap \overline{B} \neq \varnothing$, then $\tilde{B} = B \cup (A\cap\overline{B})$ is connected, since $B \subset \tilde{B}\subset \overline{B}$, and $A\cap \tilde{B} \neq \varnothing$, whence $A\cup B = A \cup \tilde{B}$ is connected. The case $\overline{A}\cap B \neq \varnothing$ is symmetric. But if $A\cap\overline{B} = \varnothing = \overline{A}\cap B$, then $A$ and $B$ are open in $A\cup B$, whence the latter is not connected.
The example of the topologists sine curve is the case the theorem handles, it is
$$\overline{\left\{(x,\sin \tfrac{1}{x}) : x > 0\right\}}.$$
