How to prove this lemma?

Question

I'm learning point set topology and read a proof for the existence of Lebesgue number in sequentially compact metric space. In the proof, the author uses such a fact :

Let $(X,\tau_{d})$ be a metric space, $A $ is a closed subset of $X$. If $p\notin A$, then $d(p,A)=\inf\{d(p,y):y\in A\}>0$.

I have tried to prove it but failed. Can anyone help me? Thanks in advance.

Answer 1

If $p \notin A$, then $p$ is an interior point of the open set $X\setminus A$ ($A$ is closed, so its complement is open), so there is some $r>0$ such that $B(p,r) \subseteq X\setminus A$. This means that if $y \in A$, we have $d(p,y) \ge r$ (or else $y \in B(p,r)\subseteq X \setminus A$ and so $y \notin A$, contradiction), and so $d(p,A) \ge r >0$.

As you can see, the lemma holds in any metric space.

Answer 2

If the infimum is $0$ then there exists $\{a_n\} \subset A$ such that $d(a_n,p) \to 0$. This means $a_n \to p$. But $A$ is closed, so the limit of any sequence in $A$ belngs to $A$. This shows $p \in A$, a contradiction.