Verfication of Proof: "$A$ is nowhere dense iff $\mathbb{R}\setminus \overline{A}$ is dense" I hope life is going well. Is possible that I get some feedback on my proof for this problem; style, accuracy, succinctness, etc..? Thank you for reading this.
Problem
Show that a set $A \subseteq \mathbb{R}$ is nowhere dense if and only if $\mathbb{R} \setminus A$ is dense.
Definitions Employed
Nowhere dense: A set $A \subseteq \mathbb{R}$ is nowhere dense if $\overline(A)^{\mathrm{0}} = \emptyset$.
(Note): $x \in A^{\mathrm{o}} \iff \exists \epsilon > 0 : \mathcal{B}_{\epsilon}(x)\subseteq A$
Dense: A set $A \subseteq \mathbb{R}$ is dense in $\mathbb{R}$ if, given any two real numbers $a, b$ it is possible to find a point $x \in A$ satisfying $a < x < b$.
My Attempt
($\Longrightarrow$) Suppose that a set $A \in \mathbb{R}$ is nowhere dense. Then for all $x \in \mathbb{R}\setminus \overline{A}$ we have $x \notin (\overline{A})^{\mathrm{o}}$, which means that $\forall \epsilon > 0$ we have $\mathcal{B}_{\epsilon}(x) \nsubseteq \overline{A}$. Now suppose for the sake of contradiction that $\mathbb{R} \setminus \overline{A}$ is not dense in $\mathbb{R}$. Then for any choice of $a, b \in \mathbb{R}$, there does not exist a point $ x\in \mathbb{R} \setminus \overline{A}$ satisfying $a < x < b$, i.e. $(a,b) \nsubseteq \mathbb{R} \setminus \overline{A}$. Since $(a,b) \nsubseteq \mathbb{R} \setminus \overline{A}$, we know it must be the case that $(a,b) \subseteq \overline{A}$. This is a contradiction since we assumed that $A$ is nowhere dense, so $\mathbb{R} \setminus \overline{A}$ must be dense.
($\Longleftarrow$) Conversely, if $\mathbb{R} \setminus \overline{A}$ is dense, then for any choice of $a, b \in \mathbb{R}$ we have $a < x < b$ for some $x \in \mathbb{R} \setminus \overline{A}$. This implies that for any choice of $a, b \in \mathbb{R}$ the non-empty open interval $(a,b) \nsubseteq (\overline{A})^{\mathrm{o}}$. Thus, $(\overline{A})^{\mathrm{o}} = \emptyset$, meaning that the set $A$ is nowhere dense.
 A: The second part is fine, but you don’t need to argue by contradiction in the first part. Suppose that $A\subseteq\Bbb R$ is nowhere dense. (Note that there is no hyphen in nowhere.) We want to show that $\Bbb R\setminus\operatorname{cl}A$ is dense, so let $a,b\in\Bbb R$ with $a<b$. Then $(a,b)\nsubseteq\operatorname{cl}A$, since $\operatorname{int}\operatorname{cl}A=\varnothing$, so there is an $x\in(a,b)\setminus\operatorname{cl}A=(a,b)\cap(\Bbb R\setminus\operatorname{cl}A)$. The points $a$ and $b$ were arbitrary (with $a<b$), so $\Bbb R\setminus\operatorname{cl}A$ is dense.
Added: On taking a closer look, I see that I ought to comment on a couple of specific points. You wrote:

Now suppose for the sake of contradiction that $\mathbb{R} \setminus \overline{A}$ is not dense in $\mathbb{R}$. Then for any choice of $a, b \in \mathbb{R}$, there does not exist a point $ x\in \mathbb{R} \setminus \overline{A}$ satisfying $a < x < b$, i.e. $(a,b) \nsubseteq \mathbb{R} \setminus \overline{A}$.

This isn’t quite right. If $\Bbb R\setminus\operatorname{cl}A$ is not dense in $\Bbb R$, there is at least one pair of real numbers $a$ and $b$ such that $a<b$ and $(a,b)\cap(\Bbb R\setminus\operatorname{cl}A)=\varnothing$. There may be other open intervals that do intersect $\Bbb R\setminus\operatorname{cl}A$ or even are contained in it, so it’s not true for just any $a$ and $b$. Note too that the non-existence of a point $x\in\Bbb R\setminus\operatorname{cl}A$ such that $a<x<b$ really does say more than just that $(a,b)\nsubseteq\Bbb R\setminus\operatorname{cl}A$: it says that $(a,b)\cap(\Bbb R\setminus\operatorname{cl}A)=\varnothing$, which is what you actually need later.

Since $(a,b) \nsubseteq \mathbb{R} \setminus \overline{A}$, we know it must be the case that $(a,b) \subseteq \overline{A}$.

This also isn’t quite right: $(a,b)\nsubseteq\Bbb R\setminus\operatorname{cl}A$ does not imply that $(a,b)\subseteq\operatorname{cl}A$. However, we actually know that there are $a$ and $b$ such that $(a,b)\cap(\Bbb R\setminus\operatorname{cl}A)=\varnothing$, which does imply that $(a,b)\subseteq\operatorname{cl}A$, and you get the desired conclusion.
A: Another "more topological" idea for $(\Rightarrow)$:
Let $A\subset \Bbb R$ nowhere dense. We want to proove that $\overline {\Bbb R\setminus \overline A}=\Bbb R$, which means that $\Bbb R\setminus \overline A$ is dense in $\Bbb R$. The first inclusion is trivial. It remains to show that $\Bbb R\subset \overline {\Bbb R\setminus \overline A}$ or that every real number is a limit point of $\Bbb R\setminus \overline A$. Let $x\in \Bbb R$ and $\epsilon>0$; if the sentence $\forall y\in [(x-\epsilon,x+\epsilon)\setminus\{x\}]:=N^*_x(\epsilon) :y\in \overline A$ was true, then immediately $N^*_x(\epsilon)\subset \overline A$, which means that $\overline A$ contains an open set, so $(\overline A)^\circ \neq \varnothing$ -a contradiction, because $A$ is nowhere dense. Consequently the true proposition is $\exists y\in N^*_x(\epsilon) :y\notin \overline A$, so $\exists y:y\in N^*_x(\epsilon)\cap(\Bbb R\setminus \overline A)$, thus the arbitrary real $x$ is a limit point of $\Bbb R\setminus \overline A$.
