Prove (or confute): distance between a point and a set 
Let $X$ be a metric space, $x\in X$ and $A\subset X$. We define: $$d(x,A)=\operatorname{inf}\{ d(x,a)\mid a\in A\}.$$ Prove or confute: There exists always $v\in\bar{A}$ such that $d(x,A)=d(x,v)$

I have some problem with this. I know that the closure of $A$ is the set of the point $x$ such that $d(x,A)=0$, but I don't know if this help nor how to use it. If $x\in A$ (or in the closure) it is clear that the equation holds. If $x\notin A$, then $d(x,A)=\operatorname{inf}\{ d(x,a)\mid a\in A\}$. This implies that, if $d(x,A)= d$, then there exists $\{a_{n}\}\subset A$ such that $d(x,a_{n})\rightarrow d$, hence there exists $v$ such that $a_{n}\rightarrow v$ and so $v\in\bar{A}$. This should work because every metric space is one-countable.
Can anyone help me? Thanks before!
 A: Here is another counterexample that I think is a bit simpler than the other answer, although it is perfectly fine. Consider the metric space $X = (-1,1) \cup \{2\}$ endowed with the metric $d(x,y) = |x-y|$. Let $x=2$ and $A=(-1,1)$. Then $d(x,A) = 1$, but since $\bar{A}=A$, there is no such $v\in \bar{A}$ with $d(x,v)=d(x,A)$.
A: It isn't true in general. While a sequence $(a_n) \in A$ exists such that $d(x, a_n) \to d$, this doesn't mean that $a_n$ converges; that would only work if $d = 0$.
It means that $a_n$ "hangs around" the ball centred at $x$, with radius $d$. If we were in a compact space (or indeed if $A$ were compact) then these point would cluster around some limit point. But, if we don't have compactness, then there's no reason to think that any sequence point gets close to any other.
Here's a counterexample. Let $X = \Bbb{N}_0 = \{0, 1, 2, \ldots\}$, and define the metric $d$ by
$$d(a, b) = \begin{cases}
0 & \text{if } a = b \\
1 + \frac{1}{a} & \text{if } 0 = b \neq a \\
1 + \frac{1}{b} & \text{if } 0 = a \neq b \\
1 & \text{if } 0 \neq a \neq b \neq 0.
\end{cases}$$
This is clearly positive definite and symmetric. It obeys the triangle inequality because $d(a, b) + d(b, c)$ is at least $2$ (unless $a = b$ or $b = c$, in which case the inequality holds vacuously), and $2$ is the maximum possible value for $d(a, c)$ (when $a = 0$ and $c = 1$, or vice-versa).
Let $x = 0$ and $A = \{1, 2, 3, \ldots\}$. Since $X$ is discrete topologically, $A$ is closed (and open), thus $\overline{A} = A$. Then, $d(x, a) = 1 + \frac{1}{a} \ge 1$ for all $a \in A$, and $d(x, a) \to 1$ as $a \to \infty$. Thus $d(x, A) = 1$.
Now, is there a point $a \in A$ such that $d(x, a) = 1$? No. Once again, we have $d(x, a) = d(0, a) = 1 + \frac{1}{a} > 1$, completing the counterexample.
