How should I prove the statement if E is a closed set, the interior of the boundary of E is an empty set? To prove this statement, i used proof by contradiction.
Suppose O$\in$int(bd(E)), the any x in O is not a boundary. Thus, O is not a subset of bdE and cannot be a subset of int(bd(E)).
However, my proof does not use that E is closed. Is my proof wrong? Where is it wrong? 
 A: Let $x \in \partial E$, where $\partial E$ is the boundary of $x$. By definition of being in the boundary, every neighborhood $U$ of $x$ intersects both $E$ and $E^c$. By definition of closure of the set $E$, $\partial E \subset E$.
Now, let $x$ be an interior point of $\partial E$. Then, there exists a neighborhood of $x$, call it $U_x$ that is contained in $\partial E$. But:
$$
U_x \subset \partial E \subset  E \implies U_x \cap E^c = \phi
$$
Therefore, $U_x$ is a neighborhood of $x$ that does not intersect $E^c$. This gives a contradiction to the fact that $x$ is a boundary point. 
Hence, when $E$ is closed, the boundary has no interior points. In the case of an open set, we cannot control the behavior of boundary points because they don't even lie in the set. Hence, $E$ being closed was necessary in this question.
A: I find it humorous the way the question says "an" empty set; as if there is more than one empty set. There is only one empty set. It is more correct to say the empty set rather than an empty set. To prove the set S is equal to the empty set, it is sufficient to only prove there does not exist any elements of S. Remember this now and also recall your definitions of boundary and interior points before we do this problem!
Let E be a subset of R
We want to show int(bd(E))=0. So we should show it's impossible for their to be an element of int(bd(E)). The easiest place to start when proving a set is empty is by going for a contradiction; so you had a great idea:
Suppose there exists an x such that x is in int(bd(E)).
Then x is an interior point of the subset bd(E) of R. This means there EXISTS a $\delta$>0 so that $N_x(\delta)\subseteq bd(E)$. Remember this. I'm labeling this small fact as (1)
Now, by definition of a delta neighborhood around x, $N_x(\delta)\textrm{, x is in }  N_x(\delta)\subseteq bd(E)$, so x is in bd(E). Thus, x is a boundary point of E. This means that, (2) for every delta greater than zero, there is an x in E and a y in R and not in E so that x and y are both in the delta neighborhood of x. 
In particular, consider the delta in fact (1). By the fact just now established as (2), there is a y not in E inside of that delta neighborhood of E, and the neighborhood is a subset of E, which means, since that y is in the neighborhood, y is in E. That is, there exists y such that y is and is not in E. This is a contradiction, so our initial supposition, that there exists an element, x, in int(bd(E)) is wrong. This necessitates the truth of its negation: that there does not exist any elements of int(bd(E)).
Thus, int(bd(E)) is empty and thus is the unique set with no elements. That is, int(bd(E))=0. 
Q.E.D
