I am learning about metric spaces and I find it very confusing. Is this a valid proof that a singleton must be closed?
If $(X,d)$ is a metric space, to show that $\{a\}$ is closed, let's show that $X \setminus \{a\}$ is open. Choose $y \in X \setminus \{a\}$ and set $\epsilon = d(a,y)$. Then since $a \not \in B(y,\epsilon)$, we have that $B(y,\epsilon) \subset X \setminus \{a\}$ so that $X \setminus \{a\}$ is open.