Suppose A has empty interior. Suppose, for a contradiction, that its closure, $A'$, is not empty. Then, WLOG, we can assume that there exists some nonempty open set $U$, and point $x$, such that $U \cap A \triangleq \{x\}$. $x$ must be on the boundary of $A$, and in the interior of $U$. As such, there remains a region of overlap between the interior of $U$ and $A$. Thus $A$ contains (at least one) nonempty open ball. Thus $A$ has nonempty interior. A contradiction.
Thank you.
Best Answer
Take $A=\mathbb Z$ for instance and $U=(-\frac 12,\frac 12)$.
$U\cap A=\{0\}$ as you stated but that's all. Your implication "there remains a region of overlap between the interior of $U$ and $A$" is true but this region is just $\{0\}$, it doesn't extend further to an open ball.