Suppose $S$ is a proper closed subset of $[0,1]$ and $\{0,1\} \subseteq S$. Show that each connected component of $[0,1]-S$ is an open interval $(a,b)$ with $a,b \in S$
Attempt: Suppose $C$ is a connected component (i.e the maximal connected subset) of $[0,1]-S$.
Then, by the property of connected components : $C$ is closed in $[0,1]-S$.
( But, since $S$ is closed in $[0,1]$, thus, $[0,1]-S$ is open in $[0,1])$
All connected components of subsets of $\mathbb R$ are intervals.
Could someone please guide me on how to move on from here. Thanks a lot.
First prove that if $U$ is open in $\mathbb R,$ then the components of $U$ are open (in $\mathbb R$): let $x\in C$, a component of $U$. Since $U$ is open, there is an open interval $x\in U_x\subseteq U.$ This implies, by the definition of component, that $U_x\subseteq C.$
Now since $[0,1]\setminus S$ is open (in $\mathbb R$), its components are open. Let $C$ be one such. Then, $C$ is an interval: for if not, then there are numbers $x<z<y$ such that $x,y\in C$ but $z\notin C$. In this case, $(-\infty, z)$ and $(z,\infty)$ are a separation of $C$.
So, $C=(a,b)$ for numbers $a,b\in \mathbb R.$ Toward a contradiction, suppose, without loss of generality, that $a\notin S.$ Then, $a\in [0,1]\setminus S$ and since this latter set is open, there is a $c<a$ such that $(c,a]\subseteq [0,1]\setminus S$. But now we have that the component $C$ is $properly$ contained in $(c,b),$ which is impossible.