The definitions given are the following:
Given a metric space $(X,d)$
A set $C \subset X $ is open iff for every $c \in C$ there exist an open ball $B(c,r) \subset C$. Where $r$ is the radius of the open ball.
A set is closed iff it's complement is open.
The boundary of a set E is the following set $\{ x \in X | \forall{r} \in \mathbb{R} \ B(x,r) \cap E \ne \emptyset \wedge B(x,r) \cap E^c \ne \emptyset\}$
With these definitions I managed to prove that the boundary is closed by showing that the complement is open, I then tried to prove it by contradiction but to no avail.
Could someone help me with a proof by contradiction using these definitions?
Suppose $\partial E$ is not closed. Then $(\partial E)^C$ is not open. So there exists $x \in (\partial E)^C$ such that no ball around $x$ is completely contained in $(\partial E)^C$. That is, every ball around $x$ intersects $\partial E$. Let $B(x,r)$ be any ball around $x$. Since $B(x,r) \cap \partial E \neq \varnothing$, there exists $y \in \partial E$ such that $y \in B(x,r)$. Then we can find a sufficiently small ball $B(y,r')$ such that $B(y,r') \subseteq B(x,r)$. Since $y \in \partial E$, $B(y,r')$ intersects both $E$ and $E^C$. That is, there exists $z_1 \in B(y,r') \cap E$ and $z_2 \in B(y,r')\cap E^C$. Because $B(y,r') \subseteq B(x,r)$, it follows that $z_1,z_2 \in B(x,r)$. Hence $B(x,r)$ intersects both $E$ and $E^C$. However, $B(x,r)$ was an arbitrary ball centered at $x$, so every ball centered at $x$ intersects $E$ and $E^C$. Then $x \in \partial E$, which is a contradiction.