Show that a set is open if and only if each point in S is an interior point. I am in a complex analysis class and have been asked to prove this.   I know I have to prove both ways so.
If a set is open then each point in $S$ is an interior point.
Proof: Let $S$ be an open set, then by definition the set contains none of its boundary points, then by definition the set $S$ contains no points that intersect both $S$ and points not in $S$.
Where do I go from here? Do I define a point $z_0 \in S$ and then say it must be an interior point because it cannot intersect both the interior and exterior of $S$ because then it would be a boundary point which contradicts the definition of open?
If each point in $S$ is an interior point then a set $S$ is open.
Let $z_0$  $\in S$
My definition for open is: A set is open if it contains none of its boundary points.
My definition for boundary points is: a  point all of whose neighborhoods contain at least one point in S and at least one point not in S. 
My definition for interior points is: a point is an interior point of the set S whenever there is some neighborhood of z that contains only points of S. 
 A: Your definition of open is quite strange, but anyways. 
Let $S$ be an open set, then it doesn't contain any boundary points. Now take $s \in S$  we want to see that $s$ is an interior point (to show that every $s\in S$ is an interior point). So by the definition you have, we want to see that there exists a neighbourhood of $s$ such that it only contains points of $S$. Let's suppose that this neighbourhood doesn't exist, then for any neighbourhood of $s$ we have that there are points which are not in $S$ which belong to the neighbourhood. But this by definition means that $s$ is a boundary point, which contradicts the asumption that $S$ is open. 
Therefore there must a exist a neighbourhood of $s$ that only contains points of $S$.
Can you do the other implication?
A: For the forward direction, you are pretty much there. There is no point in $z\in S$ with $z$ a boundary point. Well, what does it mean for a point $z$ to be a boundary point?
$$\forall \epsilon>0, z\cap S\ne \emptyset,z\cap S^{c}\ne \emptyset $$
Then if your set $S$ has no points with the above property, then for every point $z$ in your set $S$, we have some epsilon with 
$$B_{\epsilon}(z)\subseteq S$$ 
or there would be some intersection with the exterior of the set $S$, $S^c$. 
The reverse is similar. 
