The distance between a point $x$ and a set $A$ is defined as $\operatorname{dist}\left(x,A\right)=\inf\left\{d\left(x,a\right):a\in A\right\}$.
Assume that $x\notin A$. My question is, if $\operatorname{dist}\left(x,A\right)=0$, then does that mean for some $a\in A$, that $d\left(x,a\right)=0$ and that $x=a$, from the definition of a metric? But then that would mean $x\in A$? Or would it be more like that $a$ is in the open ball $B\left(x,\epsilon\right)$ for all $\epsilon>0$?
Consider $A=(0,1)$ and $x=0$. Then $d(x,A)=0$ but $d(x,a)>0$ for every $a\in A$. The best you can say is that for every $\varepsilon>0$ there is $a\in A\cap B(x,\varepsilon)$.
Note that the above condition really says that $x$ is in the closure of $A$. Therefore, if $A$ is closed, then $d(x,A)=0$ implies $x\in A$.
Let's try to understand by using an example.
let $A=(0,1)$, the open real interval from $0$ to $1$. Let $x=0$.
Hence $\operatorname{dist}(x,A)= \inf \{ d(x,a): a\in A\}=0$ since we can construct a sequence $a_n = \frac{1}{n} \in A$ such that $d(x,a_n)=\frac1{n} \to 0$
We can't have $x \in A$ since that would violate our assumption of $x \notin A$.
