Say we have $A \subset X$, given $(X, d)$ a metric space. I want to prove that the closure of $A$, that is
$$\overline{A} = \{x \in X \ | \ \forall \varepsilon > 0, \ B(x, \varepsilon) \cap A \neq \emptyset \}$$ is equivalent to $$\overline{A} = \{x \in X \ | \inf_{a \in A} d(x, a) = 0 \}.$$
What I've done is the following: by manipulating the first definition
$$\overline{A} = \{x \in X \ | \ \forall \varepsilon > 0, \ B(x, \varepsilon) \cap A \neq \emptyset \} $$ $$\ \ = \{x \in X \ | \ \forall \varepsilon > 0\ , \exists a \in A, \ d(a, x) < \varepsilon \}$$
All help is welcomed.