Let $A\subseteq\mathbb{R}$ be open. If $A\cup (0,1)$ is connected then
A must be connected.
A must have one or two component.
$A\setminus(0,1)$ has at most two component.
$A$ must be a cantor set.
Take $A=(0,1/2)\cup(1/2,1)$. Then $A\cup (0,1)=(0,1)$ is connected but $A$ is not, so $1$ is false. $2$ is also false as I can take 3 or more components. I am not sure about 3 and 4, thank you for help.