Empty interior, equivalent definitions from Munkres. The Munkres book states the following definition:

Recall that if $A$ is a subset of a space $X$ the interior of $A$ is
  defined as the union of all open sets of $X$ that are contained in
  $A$. To say that $A$ has empty interior is to say then that $A$
  contains no open set of $X$ other than the empty set. Equivalently,
  $A$ has empty interior if every point of $A$ is a limit point of the
  complement of $A$, that is, if the complement of $A$ is dense in $X$.

In such definition I don't understand the double implication
$A$ has empty interior $\Leftrightarrow$ every point of $A$ is a limit point of the complement of $A$ $\Leftrightarrow$ the complement of $A$ is dense in $X$.
I tried to formally prove the equivalence using the definition of dense set, interior etc but I just got confused.
Could you help me to understand the equivalence reported in the definition?
 A: Suppose $A$ has empty-interior. Then if $x \in A$, and $O$ is open and contains $x$, $O$ is non-empty (as it contains $x$!) so cannot be a subset of $A$ (as the interior is empty, see the remarks). So there is some point $x' \in X \setminus A$ that is in $O$. This $x'$ witnesses that $O$ intersects $X \setminus A$ in a point unequal to $x$, and as $O$ was arbitrary containing $x$, $x$ is a limit point of $X \setminus A$ (i.e. $x \in (X \setminus A)'$). So $A \subseteq (X \setminus A)'$ in this case.
On the other hand, if all $x \in A$ are a limit point of $X \setminus A$ (which Munkres denotes by $X - A$, I believe), then $A$ has a empty interior. Otherwise there would be some non-empty open set $O \subseteq A$ and any $x \in O$ is a member of $A$ which is not a limit point of $X \setminus A$, as witnessed by $O$ (as $O$ contains $x$ but no point of $X \setminus A$!). So these are equivalent.
And $X \setminus A$ is dense iff $\overline{X \setminus A} = X$ and $\overline{X \setminus A} = (X \setminus A) \cup (X \setminus A)'$. So the set is dense iff $A \subseteq (X \setminus A)'$ by simple set theory.
A: The closure of $X-A$ is equal to $\cap F$, where $F$ is family of all closed sets that have $X-A$ as a subset. The set of complements of members of $F$ is the family $G$ of all open subsets of $A.$ Hence $$Int (A)=\cup G=X-\cap F=X- \overline {X-A}.$$  $$\text {Therefore }\quad  Int (A)=\phi \iff \overline {X-A}=X.$$
