Proof of $[0,1]~\text{disconnected}\implies(0,1)~\text{disconnected}$ I want to prove the following implication
$$[0,1]~\text{disconnected}\implies(0,1)~\text{disconnected}.$$
My try:
Suppose $[0,1]=U\cup V$ with $U,V$ open, disjoint and nonempty.
Using the subspace topology of $\mathbb{R}$ we also have $U=U'\cap[0,1]$ and $V=V'\cap[0,1]$ where $U',V'$ are open in $\mathbb{R}$.
We have $(0,1)=(0,1)\cap[0,1]=(U'\cap(0,1))\cup(V'\cap(0,1))$. This is a union of open sets since $(0,1)$ is an open interval.
How can I prove that this is also a union of disjoint sets?
Will this $U\cap V=(U'\cap[0,1])\cap(V\cap[0,1])=U'\cap V'\cap[0,1]=\emptyset$ be sufficient for showing the disjoint-requirement?
Also, I do not know how to start for showing that $(0,1)$ is union of nonempty sets. 
I would appreciate any help.

You just take off the points $0,1$. 
$(0,1)=(U\setminus\{0,1\})\cup(V\setminus\{0,1\})$. Prove $U\setminus\{0,1\}$ and $V\setminus\{0,1\}$ are open in $(0,1)$ (follows almost trivially), non-empty (trivial) and disjoint (more trivial).
