A set $E$ is closed if and only if $E = E^-$ Let $E$ be a subset of a metric space $(S,d)$. I want to show that the set $E$ is closed if and only if $E = E^-$ where $E^-$ is the closure of a set $E$. 
First I assumed $E = E^-$. Then since I know $E^-$ is the intersection of all closed sets containing $E$, $E^-$ must be closed because the intersection of any closed sets is a closed set. So it follows that $E$ is closed.
Now if $E$ is closed, then $S \setminus E$ is open. That also means $E^- \setminus E$ is open since $E \subseteq E^-$ by definition. Suppose $E \subsetneq E^-$. Let $x$ be on the boundary of $E^-$. Then there does not exist an $r>0$ such that $\{s \in S \ | \ d(s, x) < r \} \subseteq E^- \setminus E$. That means $E^- \setminus E$ is not open which gives a contradiction. So it must be the case that $E^- = E$.
Could someone give me feedback on my proof? And is there a better/shorter proof?
 A: The second implication could be simpler I believe. It seems that your definition of $E^-$ is
$$
E^-=\bigcap_{C\supseteq E\ ;\ C\text{ closed}}C.
$$
If $E$ is closed, then $E$ is indeed a closed set containing $E$. Then $E$ is a member the above intersection, so $E^-\subseteq E$. Of course, $E\subseteq E^-$, so $E=E^-$. 
A: The first part of your proof is good.
If I'm understanding it correctly, the second part of your proof (where you show $E \text{ closed} \implies E = \overline{E}$)
has an error.  $\overline{E} \setminus E$ need not be open just because $E \subseteq \overline{E}$ and $S \setminus E$ is open.  For instance, if $E = [0,1]$ and $\overline{E} = [-1, 2]$, $\overline{E} \setminus E$ is not open.  Ben gives a shorter and correct proof.
A: Well I personally have learned that $\overline{E}$ (closure of $E$) is defined as $\overline{E} = E \cup E'$ where $E'$ is the set of all limit points of $E$ and that $E$ is a closed set iff $\forall \; x \in E' x \in E$ (every limit point of $E$ is a point of $E$). From this definition it's easy to see that if $\overline{E} = E$ then $E \cup E' = E$ and it follows that $E' \subset E$ so $E$ must be closed. Now if we assume $E$ is closed then $\forall \; x \in E' x \in E$ or more simply $E' \subset E$ and again from this it's easy to observe that $E \cup E' = E$.
