Show that a set $S ⊆ E$ is open if and only if $S ∩ ∂S = ∅$ 
Show that a set $S ⊆ E$ is open if and only if $S ∩ ∂S = ∅$

This is a question from my real analysis class. I started to show the forward direction by letting $S$ be open and using the definition of openness. I know that $S$ is the space inside and that the boundary of $S$ is all the points on the boundary. I can picture why their intersection is empty, but I am unsure exactly how to show this. Thank you for your help.
 A: Before reading my answer I hope that you read through the precise definitions of the objects you are considering. Elementary topology is often simply about writing down definitions in different ways until they become obvious to you. In this case all I am doing is simply writing the negations of the properties you wish to prove and arriving at a contradiction, as you will note the proofs write themselves in this case! Often topological properties are "intuitively" obvious, since they have some fundamental "geometry" behind them in some sense, but often the proofs feel rather "disconnected" from the geometry.
Assume that $S$ is open, and that $S \cap \partial S \not = \emptyset$. Now there must exist some element in $\partial S$ that is also in $S$. For all elements in $\partial S$ it follows that all neighborhoods of that element have non-empty intersections with the set $S$ and the complement of $S$. This is a contradiction since we assumed that $S$ is open, and by openness this element should have a neighborhood completely contained in $S$, but now it wouldn't. Hence $S \cap \partial S = \emptyset$.
Assume that $S \cap \partial S = \emptyset$, and that $S$ is not open. Now there must exist some element in $S$ for which every neighborhood contains some point in the complement of $S$, since otherwise $S$ would be open. Since the element is already in $S$ now every neighborhood of $S$ has a point in $S$ and its complement, so this element would be in $\partial S$ which is a contradiction since their intersection was empty. Hence $S$ must be open.
A: By definition, $\partial S = \bar S \setminus \text{int}(S)$. $S$ being open, we have $S = \text{int}(S)$. Hence..
