Open set in a metric space is union of closed sets

Show that every open set $A$ is in a metric space $(X,d)$ is the union of closed sets.

This is a question on my analysis homework. I understand that this can only be true if we consider the union of infinite closed sets. However, I am not sure what I can do. I understand to prove a set is open, then a ball centered at an arbitrary point with a radius will be completely contained in the set. But what is the radius? Is this the correct approach?

In fact, every subset of a metric space is the union of closed sets.

Hint: In a metric space singleton sets $\{x\}$ are closed.

Hint: It is equivalent to "Every closed set is the intersection of open sets". Let $C$ is a closed set. Consider $$N(C,r)=\{x\in X: d(x,y)<r \text{ for some }y\in C\}.$$ We can show that $\bigcap_{r>0} N(C,r)$ is equal to the closure of $C$