Countable subsets are disconnected

Question

I am trying to show the following: every countable subset of $\Bbb R$ with at least two points is disconnected.

My attempt: let $D$ be such subset. Then take $a \in D$ and define $A=\{ a\}$ and $B = D - \{ a\}$. I want to use the result that a set is disconnected if and only if there exists mutually separated sets whose union is the set. I tried to show that $A$ and $B$ are mutually separated sets using $\overline{A} = \{ a \}$ and $\overline{B}$ but $\overline{B} = \overline{D - \{a \}} = D - \{a\}^{\circ} = D$. We have $\overline{A} \cap B = \emptyset$ but $A \cap \overline{B} = \{a\} \neq \emptyset$.

I'm not seeing another line of attack.

Answer 1

Your approach seems to imply that every point in a countable set is isolated. Namely that a countable is synonymous with "discrete" in the case of the real numbers.

This is not true, of course. The rational numbers are countable but not discrete.

Instead, let me give you the following hint:

If $D$ is a countable subset of $\Bbb R$, show that there is some $x\in\Bbb R$ such that $x\notin D$ and neither $D\cap(-\infty,x)$ nor $D\cap(x,\infty)$ are empty.

Answer 2

Note that a subset of $\mathbb{R}$ is connected iff it's an interval.

Now let $D\subset\mathbb{R}$ be atmost countable with at least two elements, supposing it to be connected will exhibit a contradiction, since an interval has uncountably infinite elements (except $\space \{a\}\space \forall a\in \mathbb{R}$)

Warning:$(a,b)_Q$ for $a,b\in\mathbb{Q}$ denoting the induced topology on $\mathbb{Q}\cap(a,b)$ by $\mathbb{R}$ is not an interval of $\mathbb{R}$, in most books, it's not even considered to be an interval at all!