$(X,\tau)$ be a topological space and $A\subseteq X$.
\begin{equation}A\text{ nowhere dense in }X \iff \overline{A}\text{ has empty interior.}\; \; \; \;\;(1)\end{equation}
I know that this means there exists no $E \in \tau$ such that $A$ is dense in $E$ w.r.t to the subset topology $\tau_{E}$. This definition seems more intuitive than the original one above.
One issue I am battling with is the NEED for the closure $ \overline{A}$ to have empty interior. Why not define $A$ as nowhere dense in $X$ iff $A$ has empty interior.
My question is basically, what would change if in $(1)$ $A$ was required to have empty interior and not $\overline{A}$?
I may be missing something elementary here.
The intention behind the concept of a set being dense is to say when a subset $A$ of a set $X$ has the property that each point in $x\in X$ is arbitrarily close to some point $a$ in $A$ (or is in $A$ itself), see e.g. here.
If you translate this into a statement about sequences in $A$ approximating elements of $X$ you will see that a set $A$ is dense in $X$ if $\bar A = X$. (This explanation is correct when you can describe the closure of a set using convergent sequences, which is true in metric spaces. Similar ideas can be applied in the general case. That's not the focus here).
A well known example of a dense subset is the set of the rational numbers, which is known to be a dense subset of the real numbers.
So if you want to define the opposite of an (everywhere) dense set, you take the defining property (the closure being everything) and ask that it, instead of being everything, does not contain 'too many' points. One way of saying that a set does not contain too many points, which is specifically suited to topological spaces is to ask for an empty interior. So you arrive at: $(\bar A)° = \emptyset$
Just asking for $A$ to have empty interior is too strict. Many sets made up of distinct points have empty interior (like $\mathbb{Q} $ in $\mathbb{R}$) but are, regardless of this, dense.