Boundary Points and Metric space Definition:The boundary of a subset of a metric space X is defined to be the set $\partial{E}$ $=$ $\bar{E} \cap \overline{X\setminus E}$ 
Definition: A subset E of X is closed if it is equal to its closure, $\bar{E}$.
Theorem: Let C be a subset of a metric space X. C is closed iff $C^c$ is open. 
Definition: A subset of a metric space X is open if for each point in the space there exists a ball contained within the space
Show that if $E \cap \partial{E}$ $=$ $\emptyset$ then $E$ is open.
Proof:
$E\cap \partial{E}$ being empty means that $ E\subseteq (\bar{E}^c \cup \overline{X\setminus E}^c)$. Since $E \subseteq \bar{E}$ it follows that $E \subseteq \overline{X\setminus E}^c$ which implies that $E \cap \overline{X\setminus E}$ is empty. Since every subset is a subset of its closure, it follows that $X\setminus E$ $=$ $\overline{X\setminus E}$ and so $X\setminus E$ is closed, and therefore $E$ is open.
Is the proof correct? I would really love feedback.
 A: After William Elliot's feedback on your proof and this comment of yours, I don't think there is much that needs to be clarified. Still if you have anything specific regarding your proof to ask me, I welcome you to come here. 
In any case, let me try to write a proof that I believe is in line with your attempt.

\begin{align*}E\cap \partial{E}=\emptyset&\implies E\cap(\overline{E}\cap \overline{X\setminus E})=\emptyset\\&\implies (E\cap\overline{E})\cap \overline{X\setminus E}=\emptyset\\&\implies E\cap \overline{X\setminus E}=\emptyset\\&\implies \overline{X\setminus E}\subseteq X\setminus E\\&\implies \overline{X\setminus E}=X\setminus E\end{align*}This shows that $X\setminus E$ is closed and hence $E$ is open.

A: In any  topological space $X$ and any $E\subset X,$ the 3 sets $int(E),\, int(X\setminus E),\,  \partial E)$ are pair-wise disjoint and their union is $X.$
So if $E\cap \partial E=\emptyset$  then $$E=E\cap X=E\cap (\,int (E) \cup int (X\setminus E)\cup  \partial E\,)=$$ $$=(E\cap int E)\,\cup\, (E\cap int (X\setminus E))\,\cup\, (E\cap \partial E)\subset$$ $$\subset (E\cap int(E)\,\cup \,( E\cap (X\setminus E)\,\cup\, (E\cap \partial E)=$$ $$=int (E)\,\cup \, ( \emptyset)\,\cup \,(\emptyset)=$$ $$=int (E)\subset E$$ so $E=int(E).$
OR, from the first sentence above, for any $E\subset X$ we have $int(E)\subset E\subset \overline E=int(E)\cup \partial E.$
So if $E\cap \partial E=\emptyset$ then $$E=E\cap \overline E=E\cap (int (E) \cup \partial E)=$$ $$=(E\cap int (E))\,\cup \,(E\cap \partial E)=$$ $$=(E\cap int (E))\cup(\emptyset)=$$ $$=int(E)\subset E$$ so $E=int(E).$
