Prove that if $X$ and it's closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected. Can anyone give me a proof of the statement that if $X$ and its closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected?
Thank you. 
 A: In case you are using the "you can't write it as a union of disjoint open sets" definition, like I grew up with:
Let $O_1$ and $O_2$ be open sets of the universal set whose union contains $Y$ and such that $(O_1\cap Y)\cap(O_2\cap Y)=\emptyset$. 
Since their union also contains $X$, which is connected, one of them, say $O_1$ is such that $X\cap O_1=\emptyset$ (and $X\subseteq O_2$). 
But then by definition of the closure, $\overline{X}\cap O_1=\emptyset$, and hence $Y\cap O_1=\emptyset$. So, $Y\subset O_2$.
A: If $Y$ is not connected, $Y$ is union of nonempty seperated set $A,B$. Since $X$ is connected, Then $X\subset A$ or $X\subset B$.
If $X\subset A$, then $\overline{X}\subset \overline{A}$. So $Y\subset \overline{A}$ But since $A$, $B$ is seperated, $\overline{A}\cap B$ is empty and $B\subset \overline{A}$, so $B=\varnothing$, which derive a contradiction.
A: Let $g : Y \to \{0,1\}$ be a continuous function. Then $g_{|X} : X \to \{0,1\}$ is also continuous, and because $X$ is connected, $g$ is constant on $X$; because $X$ is dense in $Y$, you deduce that $g$ is constant on $Y$.
Therefore, every continuous function from $Y$ to $\{0,1\}$ is constant: $Y$ is connected.
