Showing that S is not both open and closed I'm working on a midterm practice exam and needed some help finishing off the proof. For those who are interested this is Undergraduate Topology by Robert H. Kasriel (Chapter 44)

Let $S \subseteq \mathbb{R}$ be such that $S \neq \emptyset$ and $S \neq \mathbb{R}$. Show that $S$ is not both open and closed (with respect to the Euclidean metric)

I think I am to use a proof by contradiction but am lost after I've assumed it to be open and closed and how to end it.
Here is my attempt so far (to note I like to use bullet points in my proofs as some form of sequential order helps me visualize my thinking):
Proof.

*

*Suppose that the subset $S$ (such that $S \neq \emptyset$ and $S \neq \mathbb{R}$) is both open and closed

*Then since $S$ is closed, its complement $S^{c} = \mathbb{R} \backslash S$ will be an open set

 A: You'll need to use the completeness property of $\Bbb R$.  Note that this statement is false for, say, $\Bbb Q$; let $S= \{ r \in \Bbb Q \mid r \gt \sqrt 2 \}$.
My approach would be as follows:  If $S$ is open and bounded below, show that $\inf(S) \notin S$ (using that $S$ is open) and no neighborhood of $\inf(S)$ can be contained in the complement of $S$ (using the definition of $\inf(S)$), so that complement can't be open.  The same argument works (using $\sup (S)$) if $S$ is bounded above.
If $S$ is not bounded above or below, a similar argument still works.  Choose $y \notin S$ and then choose $x= \sup \{ z \geq y \mid \forall w \in [y, z]~(w \notin S)\}$.  Prove $x$ exists, $x \notin S$ (using that $S$ is open), and every neighborhood of $x$ intersects $S$, so the complement of $S$ again can't be open.
A: I will assume that you know the real numbers are connected. If $\emptyset\subset S\subset\mathbb{R}$, is both open and closed, then $\mathbb{R}\setminus S$ is also open and closed.  As $S$ and $\mathbb{R}\setminus S$ are non-empty, disjoint, open sets whose union is all of $\mathbb{R}$, then you can conclude that $\mathbb{R}$ is disconnected, which is a contradiction.
(This generalizes to any topological space, that is, a topological space $X$ is connected if and only if the only sets which are both closed and open are $\emptyset$ and $X$)
