$A$ is open in metric space $X$ if and only if $X \setminus A$ is closed We were given a theorem in class:

$A$ is open in a metric space $X \Longleftrightarrow X \setminus A$ is closed.

But the proof was omitted, so I am trying it for myself.
We call $S$ closed in $X$ if the set of limit points $S^\prime = \{x \in X: \forall \varepsilon > 0, B_{\varepsilon}(x) \cap (S \setminus \{x\}) \ne \emptyset \}$ is contained in $S.$ (i.e. $S^\prime \subseteq S).$ 
We call $S$ open in $X$ if $S$ is contained in the set of interior points $S^\circ = \{x \in X:\exists B_{\varepsilon}(x) \subseteq S \}.$ (i.e. $S \subseteq S^\circ)$
Work so far:
$(\Rightarrow):$ Set $B := X \setminus A.$ Take $x \in B^\prime$ and assume $x \not\in B$ (to the contrary that $B$ is closed). By assumption, $A$ is open, so if we take $x \in A,$ then there exists $\varepsilon > 0$ such that $B_{\varepsilon}(x) \subseteq A.$ Thus $B \cap B_{\varepsilon}(x) = \emptyset,$ which contradicts that $x \in B^\prime,$ so $B$ must be closed.
$(\Leftarrow):$ How do we prove this direction?
 A: Suppose that $X \setminus A$ is closed, $x\in A$ suppose for every integer $n>0,$ $B_{1/n}(x)$ is not contained in $A$ and let $x_n \in B_{1/n}(x) \cap (X \setminus A)$, the limit of $x_n$ is $x,$ since $X \setminus A$ closed. Thus $x \in X \setminus A$ contradiction since $A\cap (X \setminus A)$ is empty.
A: Suppose the complement of $A$ denoted by $A'$ is closed. We claim that $(A')'$ is open. Pick some $p\in (A')'$. If there does not exist some r-neighborhood around it such that $d(p,q)<r \implies q\in (A')'$, then for each $r=\frac{1}{n}$, for $n=1, 2, 3,...$ there exists a point $q_n\in A'$ such that $d(p, q_n)<\frac{1}{n}$. This sequence of $q_n$'s converges to a limit in $(A')'$. But this contradicts the fact that $A'$ is closed, therefore the complement must be open. 
A: Proof without sequences (which are not needed here). 
Suppose $X \setminus A$ is closed. We want to show that $A$ is open. So pick $x \in A$. Suppose (for a contradiction) that $x$ is not an interior point of $A$. Then for every $r > 0$, $B(x,r) \nsubseteq A$, so for every $r > 0$, $B(x,r) \cap (X \setminus A) \neq \emptyset$. The points in which the balls around $x$ intersect $X \setminus A$ are by definition points unequal to $x$, as $x \in A$. So $x \in (X\setminus A)' \subseteq X \setminus A$, as the latter set is closed by assumption. But then $x \notin A$, contradiction. So $x$ is an interior point of $A$, and as $x$ was arbitrary, $A$ is open.  
