Nowhere dense iff not dense in any nonempty open subset I use the following definitions:

The set $A\subset X$ is said to be dense in $X$ if $\bar A=X$.
The set $A\subset X$ is said to be nowhere dense in $X$ if $(\bar A)^\circ=\emptyset$.

Yesterday I read a post saying that a set $A$ is nowhere dense in $X$ if and only if $A$ is not dense in any nonempty open subset of $X$.
If $A$ is nowhere dense, by the definition I follow, $(\bar A)^\circ=\emptyset$. It implies that $\bar A$ is not a nonempty open set. That is, $\bar A≠G$ where $G≠\emptyset$ is open. So $A$ is not dense in any nonempty open subset of $X$. I'm confused about the converse part :
Take $A=[0,1]$. Then $(\bar A)^\circ=(0,1)≠\emptyset$. So $A$ isn't nowhere dense. Then $A$ (according to the post) must be dense in some nonempty open subset of $X$. What is that set? $\bar A$ is not an open set. How can the statement be true?
 A: Regarding your confusion for the converse part, $A$ is dense in the non-empty open set $G = (0,1)$. This is because by the definition of denseness, $A$ is dense in $G$ iff $A \cap G$ is dense in $G$. That is, iff $G \cap (\overline{A \cap G}) = G$. Here, $A \cap G = G$, and $\overline{A \cap G} = A$, so $G \cap (\overline{A \cap G}) = G$.

Additionally, the proof you have given for the forward implication is slightly incorrect. You say that $(\bar{A})^0 = \emptyset \implies \bar{A}$ is not a nonempty open set. But this is assuming too much; the correct implication is: $(\bar{A})^0 = \emptyset \implies \bar{A}$ does not contain any nonempty open set.
To correct the proof, one could work as follows. Let $(\bar{A})^0 = \emptyset$. Then, $\bar{A}$ does not contain any nonempty open subset of $X$. Let $G$ be a nonempty open subset of $X$. If $A$ is dense in $G$, then $$
G \cap (\overline{A \cap G}) = G.
$$
But,
$$
A \cap G \subset A \implies \overline{A \cap G} \subset \bar{A}.
$$
Therefore, $G \cap (\overline{A \cap G}) \subset G \cap \bar{A} \subset G$. Hence,
$$
G \cap (\overline{A \cap G}) = G \implies G \cap \bar{A} = G \implies G \subset \bar{A}.
$$ 
But this contradicts that $\bar{A}$ does not contain any nonempty open subset of $X$. Hence, $A$ is not dense in $G$.

Further comments: Your proof breaks down at the point where you conclude from $\bar{A}$ not being a (nonempty) open set to the fact that $A$ is not dense in any nonempty open subset of $X$. In general this is not true.
To see why, we can take the same sets you considered for the reverse implication. Let $X = \mathbb{R}$, $A = [0,1]$. Then $\bar{A} = A$, which is not a nonempty open set. But $A$ is still dense in the set $G = (0,1)$. So, just having that $\bar{A}$ is not a nonempty open set does not imply that $A$ is not dense in any nonempty subset of $X$.
