Proof verification: Any countable subset of $\Bbb R$ is disconnected In an exercise I'm asked to prove the following:

Prove that every countable subset of $\Bbb R$ with more than one point is disconnected.

I did My proof but there's one set on the proof that I'm not sure how to prove. This is my proof:
Let $S\subset \Bbb R$ such that $S$ is countable. This means that we can list the elements of the set $S$ as following:
$$S = \{ s_1,...,s_n\}$$
Where $n$ can be a natural number if the set is finite, or $n$ can go infinitely high if the set $S$ is countably infinite.
Now let's define the following interval:
$$I = (\inf S,\sup S)$$
Because $\text{card } I > \text{card } S$ we can assume that $S \subset I$. So there are element of $I$ that do not belong to $S$. So let $y\in I\setminus S$. This means that there exists $s_i,s_j\in S$ such that:
$$s_i < y < s_j\ \ \ (1)$$
We have that $s_i,s_j\in S$ but $y \notin S$. This means that $S$ is not an interval and therefore it's not connected.

First of all: Is this proof correct? I'm not quite sure how to prove statement (1). How can I prove that such $s_i,s_j \in S$ do in fact exists?
 A: The idea is good, but there are a few problems. For instance, the set $S$ may not have supremum or infimum. That's not a really serious problem, since you can use $\infty$ instead of $\sup S$ and $-\infty$ instead of $\inf S$ then.
However, you can simply take $s_1,s_2\in S$ with $s_1<s_2$ and consider the interval $(s_1,s_2)$. Since its cardinal is greater than the cardinal of $S\cap(s_1,s_2)$, there is some $y\in(s_1,s_2)$ such that $y\notin S$. So, $S$ is not an interval and therefore it is not connected.
A: The statement of the OP should be
If $S\subset\mathbb{R}$ is coutable and has more than one point then $S$ is disconnected.
A proof along the line of the OP's reasoning is as follows:
Let $s_1,s_2\in S$ such that $s_1<s_2$. Let $r\in \mathbb(s_1,s_2)\setminus S$. Such point exists since $(s_1,s_2)$ is uncountable. Then $A=S\cap (-\infty,r)$ is nonempty open and closed subset of $S$; similarly $B\cap(r,\infty)$ is nonempty open and closed subset of $S$.
Also $S=A\cup B$.
A: Pick $s_1 < s_2$ from $S$. Then $[s_1,s_2]$ contains some $t \in S^c$. Then $(-\infty,t), (t,\infty)$ are open disjoint sets that intersect $S$ and  whose union contains $S$ and so $S$ is disconnected.
A: Alternatively, the connected subsets of $\Bbb{R}$ are intervals and non-singelton intervals are uncountable.
