I am writing down a sketch of the proof to the following problem in words and would appreciate your 2cent!
Given an arbitrary metric space $(X,δ)$, show that if a set $Y \subset X$ is open, then its complement is closed.
Sketch of the proof (in words):
(1) Given the arbitrary metric space, suppose $Y\subset X$ is open.
(2) Suppose $Y^c$, the complement of $Y$, is not closed.
(3) By (2), there exists a limit point of $Y^c$ that is not contained in $Y^c$. Denote this point $y^*$.
(4) $y^*$ is in $Y$, and there exists a sequence in $Y^c$ that converges to $y^*$.
(5) By (4), $\forall\epsilon>0$, $\exists y_c\in Y^c$, $y_c\in B_\epsilon(y^*)$.
(6) But we assumed Y is open.
This is a contradiction. It must be that $Y^c$ is closed. QED.