Show that For $A$ closed in a metric space $(,)$ and $a$ isn't in A, $(,A):=inf[(a,x) : x \in A]$ > 0

Question

I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.

Can I just say:

Let $a\in X, a \notin A$ and $y \in A$. Then $\forall y \in A$ , $dist(a,x)>0 \implies inf[(a,x)| x \in A] > 0$

Answer 1

Assume $d(A, a) =0$.
By definition of the infimum, there's an element $x_n\in A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.
But then $x_n\to a$ follows, and if $A$ is closed, it implies $a\in A$.