I am to prove that cl(E) is closed. The only two definitions for closure that I am allowed to work with are that cl(E) $ = \{x \in X: \forall $ neighborhoods $U$ of $ x, U\cap E \neq \emptyset\}$ and cl(E) = $X \setminus(X\setminus E)^{\circ}$, where $x \in E^{\circ}$ means there exists a neighborhood $U$ of x such that $x \in U \subset E$.
I cannot use limits, boundaries, open balls, cl(E) $= E \cup E$', or the fact that cl(E) is the smallest closed set containing E.
So I want to show cl(E) is closed by showing $X \setminus$cl(E) is open. Here's what I have so far...
Proof sketch: Let $y \in X \setminus$cl(E). Since $y \notin$ cl(E), there exists a neighborhood $U$ of y such that $U\cap E = \emptyset$. Clearly $y \in U$ and since $U\cap E = \emptyset$, then y $\in X\setminus E$. At this point I'm thinking I can say that by the above definition of interior, $y \in (X \setminus E)^{\circ}$, because $U \subset (X \setminus E)$. I am not allowed to use the fact that the interior of a set is open, so now I am feeling stuck.
Any hints would be greatly appreciated, but please do not post full answers.
Let $x \in cl(E)^c$. Then there exists $U \subseteq X$ an open neighborhood of $x$ such that $E \cap U = \emptyset$. Try to show that $U \cap cl(E) = \emptyset$, and conclude that $cl(E)^c$ is open.
EDIT: I edited away a lot of the answer since OP does not want full answers.