Prove that $X\setminus A$ is nowhere dense iff for every dense subset $B$ of $X$ we have $\overline{A\cap B}=X$. HELPFUL DEFINITION: Let (X,d) be a metric space. A subset B of X is said to be dense in X if $\overline B$ = X. For example, both $\mathbb{Q}$ and $\mathbb{R}$ \ $\mathbb{Q}$ are dense in $\mathbb{R}$. A subset B of X is said to be nowhere dense in X if Int($\bar{\text{B}}$) = ∅ (meaning that B is nowhere dense in X if and only if the interior of the closure of B is empty).
QUESTION: Let (X,d) be a metric space. Let A be a subset of X. Prove that X\A is nowhere dense if and only if for every dense subset B of X we have $\overline{A \cap B} = X$.
 A: I'll list some lemmas to consider. (While they are only stated for metric spaces, each are true in general topological spaces). If you have trouble proving or using them, please let me know.
Since your question involves complements, interiors, and closures, it's good to know how they interact with one another.

Lemma 1. Suppose $A$ is a subset of a metric space $(X,d)$. Then
  $$
\overline{A^c}=(\operatorname{int}(A))^c \quad\text{and}\quad \operatorname{int}(A^c)=(\overline{A})^c,
$$
  where $A^c$ denotes the complement $X\setminus A$.

The next lemma is a very useful "definition" of closure. It makes proving Lemma 3 easy.

Lemma 2. If $A$ is a subset of a metric space $(X,d)$, then $x\in\overline{A}$ iff for every open subset $U$ containing $x$, we have $A\cap U\ne\emptyset$.

Here's a nice result regarding the intersection of dense sets.

Lemma 3. Suppose $A$ and $B$ are dense subsets of a metric space $(X,d)$. If $A$ is open, then $A\cap B$ is dense in $X$.

A short proof of your problem follows by using lemmas 1 and 3, as well as the simple facts that supersets of dense sets are dense and that the whole space $X$ is dense.
A: (0). If $X=\phi$ then $X$ \ $A$ is $not$ nowhere dense and any $A\cap B \subset X$ satisfies $\overline {A\cap B}=X.$ 
(1). If $X\ne \phi$ and $X$ \ $A$ is  $not$  nowhere dense let $C\ne \phi$ be open with $C\subset \overline {X\setminus A}.$  Let $B=(X \setminus A)\cup (X \setminus  C).$ Then $B$ is dense in $X .$ 
Now $C$ is open and disjoint from $A\cap B$ so $C$ is disjoint from $\overline {A\cap B}.$ Since $C\ne \phi$  we have $\overline {A\cap B}\ne X.$
(2). If $X$ \ $A$ is nowhere dense let $C$ be any non-empty open set, and let $D$ be open and not empty with $D\subset C$ and $D\cap \overline {X \setminus A}=\phi.$ So $D\subset A.$  
Now if $B$ is any dense subset of $X$ then   $$\phi \ne D\cap B= (D\cap A)\cap B\subset (C\cap A)\cap B)=C\cap (A\cap B).$$   So $A\cap B$ has non-empty intersection with any non-empty open $C.$ Therefore $A\cap B$ is dense in $X.$ That is, $\overline {A\cap B}=X.$
Remark: In (2) we could take $D=C\setminus \overline {X\setminus A}$.
Remark: In general, if $E$ is open and $B$ is dense then $\overline E=\overline {E\cap B}.$ So we could do (2), if $X$ \ $A$ is nowhere dense, by letting  $E=X \setminus \overline {X\setminus A}.$ Then  $E$  is dense and open and $E\subset A.$ Since $E$ is open, if $B$ is dense then $X=\overline {E}=\overline {E\cap B}\subset \overline {A\cap B}\subset X.$
