Question
Consider open set of real numbers $A$ such that $A\ne\emptyset$ and $A\ne \mathbb{R}$. Denote by $Bd(A)$ a set of boundary points of $A$. Is it true that $Bd(A)\ne\emptyset$?
It seems that according to the answer to this question the statement is true. However it uses the fact that $\emptyset$ and $\mathbb{R}$ are the only sets of real numbers that are open and closed simulteneously. However I whant to use abovementioned statement as a lemma to proof that $\emptyset$ and $\mathbb{R}$ are the only clopen sets. Therefore any alternative proof or comments on mistakes on my understanding of this answers are hygly appreciated.
Cosiderations
I think that the statement is true since intuition suggests that boundary point will be supremum (or infinum) of some interior neighborhood of point from $A$ and this supremum (or infinum) will be a point outside of $A$. For example for an open set $A=(-\infty,1)$ we may choose some interior neighborhood $N_{0.1}(0.9)=(0.8,1)$. Then it's supremum is $\sup(N)=1\notin A$ that is also a boundary point of $A$ so $Bd(A)\ne\emptyset$.
Proof
My attempt of direct proof in general case is as follows.
Since $A\ne\emptyset$ there is $x\in \mathbb{R}$ such that $x\in A$.
Consider a set: $$S = \{\varepsilon>0:N_{\varepsilon}(x)\subset A\},$$
where $N_{\varepsilon}(x)=(x - \varepsilon, x + \varepsilon)$ is an open interval.
Since $A\ne \mathbb{R}$ there is $y\in\overline{A}$ where $\overline{A}=\mathbb{R}-A=\{t\in \mathbb{R}:t\notin A\}$.
Consider $\varepsilon^{*}>|x-y|$ then:
$$[x - |x-y|, x + |x-y|]\subset(x-\varepsilon^{*},x+\varepsilon^{*})=N_{\varepsilon^{*}}(x),$$
where:
$$[x - |x-y|, x + |x-y|]=\begin{cases} [y,2x-y]\text{, if }x>y\\ [2x-y,y]\text{, if }x<y \end{cases}.$$
Hence $y\in N_{\varepsilon^{*}}(x)$ so $N_{\varepsilon^{*}}(x)\not\subset A$ implying that $\varepsilon^{*}$ is an upper bound of $S$. Consequently by the completeness axiom $S$ has supremum $\sup(S)$. The idea is to show that $x-\sup(S)$ or $x+\sup(S)$ is a boundary point of $A$.
There are two possible cases.
Case 1. Suppose $\sup(S)\in S$. Then $N_{\sup(S)}(x)=(x-\sup(S),x+\sup(S))\subset A$. Since $\sup(S)$ is supremum of $S$ for any $\varepsilon > 0$ we have $\sup(S)+\varepsilon\notin S$ so:
$$N_{\sup(S)+\varepsilon}(x)=(x-\sup(S)-\varepsilon,x+\sup(S)+\varepsilon)\not\subset A.$$
Combining this result with the fact that $N_{\sup(S)}(x)\subset A$ we get: $$(x-\sup(S)-\varepsilon, x-\sup(S)]\not\subset A,$$ or $$[x+\sup(S), x+\sup(S)+\varepsilon)\not\subset A.$$
Consider any $\varepsilon$ such that the latter holds (for the former case arguments are similar so they are omitted for brevity), then:
$$[x+\sup(S),x+\sup(S)+\varepsilon)\subset(x+\sup(S)-\varepsilon,x+\sup(S)+\varepsilon)=N_{\varepsilon}(x+\sup(S)).$$
Consequently $N_{\varepsilon}(x+\sup(S))\cap \overline{A}\ne\emptyset$. Let's denote it result 1).
Further observe that:
$$N_{\varepsilon}(x+\sup(S))\cap N_{\sup(S)}(x)=(x+\sup(S)-\varepsilon, x+\sup(S))\ne\emptyset\text{, if }\sup(S)\geq\varepsilon.$$
$$N_{\varepsilon}(x+\sup(S))\cap N_{\sup(S)}(x)=(x-\min(\sup(S), \varepsilon - \sup(S)), x+\sup(S))\ne\emptyset\text{, if }\sup(S)<\varepsilon.$$
Since $N_{\sup(S)}(x)\subset A$ we get that $N_{\varepsilon}(x+\sup(S))\cap N_{\sup(S)}(x)\ne\emptyset$ implies $N_{\varepsilon}(x+\sup(S))\cap A\ne\emptyset$. Let's denote it result 2).
By definition of boundary point results 1) and 2) imply that $x+\sup(S)\in Bd(A)$ so $Bd(A)\ne\emptyset$.
Case 2. Suppose $\sup(S)\notin S$. Then $N_{\sup(S)}(x)=(x-\sup(S),x+\sup(S))\not\subset A$. Since $x\in A$ we have $(x-\sup(S),x)\not\subset A$ or $(x,x+\sup(S))\not\subset A$. Without loss of generality suppose latter. Then since $(x,x+\sup(S))\not\subset A$ there is $z\in (x,x+\sup(S))$ such that $z\notin A$ so $(x,z]\not\subset A$. Since $\mathbb{R}$ is dense there is $z^{*}\in R$ such that $z<z^{*}<\sup(S)$ so $(x,z]\subset(x,z^{*})$ hence $(x,z^{*})\not\subset A$. Consider $\varepsilon = z^{*} - x$. Since $z^{*}<x+\sup(S)$ we get $\varepsilon+x<x+\sup(S)$. Hence there is $\varepsilon < \sup(S)$ such that $N_{\varepsilon}(x)\not\subset A$ so $\varepsilon$ is upper bound of $S$ that is contradiction to the fact that $\sup(S)$ is supremum of $S$. Consequently this case is impossible.
Q.E.D.
P.S.
Will be extremely grateful for any comments on my proof as well as for disproof or alternative proof of the statement!
Edit
Following comment of Anne Bauval (thank you very much for the response!) I have removed "without loss of generality" statement.