If $A\subseteq\Bbb R$ is nonempty with $|A|\ge 2$, then $A$ totally disconnected $\iff A^\circ=\emptyset$.

Question

In the course of working on an exercise, I came up with the claim given in the title. Just looking for verification.

$\underline{\text{Claim: } A\text{ is totally disconnected}\iff A^\circ=\emptyset.}$

$\underline{\implies}$

Let $x\in A$, $\epsilon>0$ and $U=(x-\epsilon, x+\epsilon)$. Since $U\cap A$ is not connected, $U\cap (\Bbb R\setminus A)\ne\emptyset$ and therefore, $x\notin A^\circ$.

$\underline{\impliedby}$

Let $x,y\in A$ with $x<y$. Suppose there is a connected set $B\subseteq A$ such that $x,y\in B$. Since, $(x,y)$ is connected, $(x,y)\subseteq B\subseteq A$. This contradicts that $A^\circ=\emptyset$.

Answer 1

I would modify the proof a bit. In the first sentence you could write

Let $x∈A, ϵ>0$ and $U=(x−ϵ,x+ϵ)$. Since $U∩A$ $\color{red}{\text{is a singleton}}$ or is not connected, $U∩(ℝ∖A)≠∅$ and therefore, $x∉A^∘$.

The other direction is fine.

I think you don't need the hypothesis that $|A|\ge 2$. If $A$ is a single point, then it's totally disconnected since the components are singletons in that case.