Proving that a set $A$ is dense in $M$ iff $A^c$ has empty interior 
Prove that a set $A$ is dense in a metric space $(M,d)$ iff $A^c$ has empty interior.

Attempt:
I think I proved the converse correctly, but I'm not sure how to start the forward direction.
$(\Leftarrow)$
Assume $A^c$ has empty interior. Show that $M=\operatorname{cl}(A)$, where $\operatorname{cl}(A)$ denotes the closure of $A$. We also note that $\operatorname{cl}(A)=(\operatorname{int}(A^c))^c=\emptyset^c=M$, where $\operatorname{int}(A^c)$ denotes the set of interior points of $A^c$.
$\blacksquare$
Any help would be appreciated. Thanks.
 A: This stems from the more general fact (see this question) that given $A \subseteq X$ $$\operatorname{Int} (A) = X \setminus \overline{ X \setminus A },$$ or, equivalently, $\operatorname{Int}( X \setminus A ) = X \setminus \overline{A}$. (Note that this is valid in all topological spaces, not just metric spaces.)
So $A$ is dense if and only if $X = \overline{A}$ if and only if $\varnothing = X \setminus X = X \setminus \overline{A} = \operatorname{Int} ( X \setminus A )$.
A: Just recall that in any topological space $X$ the closure of the complement is equal to the complement of the interior. Formally $(E^\circ)^c=\overline{(E^c)}$ for any $E\subseteq X$. Take $E^\circ=\emptyset$ for one implication. Assume $E^c$ to be dense to get the other implication.
A: $\implies$  
Suppose that $A$ is dense in $M$. Then $cl(A)=M$. If $M$ is empty then the result is trivial. So assume $M\ne\phi$. The $cl(A)\ne \phi$. Then clearly, $A\ne \phi$ otherwise $A=\phi=cl(\phi)\ne M$. Thus $A\ne \phi \implies A^c=\phi$ and so its interior. $\blacksquare$  
Converse what you have proved seems to be right.
