The question has already been answered over here
but I was wondering if my approach is correct or not.
My attempt: Let $S$ be a nonempty proper subset of $\mathbb{R}$ and furthermore assume $S$ is both open and closed in $\mathbb{R}$. Since $S$ is open, $S$ is the union of countable family of pairwise disjoint open intervals. Let $(a,b) \subset S$ be the one of such pairwise disjoint open interval.
We claim that $S$ does not contain the point $a$. If it were the case that $a\in S$ then $a\in (c,d)$ where $(c,d) \subset S$ is one of open interval from the family of disjoint open interval. Consider the point $x=\frac{a+ \min \{ b , d \} }{2}$. Then $x \in (a,b) \cap (c,d)$. But this is a contradiction to the fact that $(a,b) , (c,d)$ were from the family of pairwise disjoint open intervals. Hence, $a \not\in S$.
But $a$ is an accumulation point of $S$ and since $S$ is closed, it contains all its accumulation points, thus, $a\in S$.
We have that $a\in S$ and $a \not \in S$, a contradiction! We are done.
Is this proof correct?