In a metric space, the closed ball $\bar{B}(x,r)$ is not always the closure of the open ball $B(x,r)$ The question reads:

Let (X,d) be a metric space.  For $x \in\ X$ and $r>0$ the closed ball of radius $r$ centered at $x$ is the set $\bar{B}(x,r)$ defined by $\bar{B}(x,r) = \{y \in X: d(x,y) \leq r\}$.
Show that the set $\bar{B}(x,r)$ is indeed closed in $X$, but that this set need not be equal to $\overline{B(x,r)}$, the closure of $B(x,r)$ in $X$.

I've separated the problem into two parts, showing $\bar{B}(x,r)$ is closed, and then showing that it is not necessarily equal to the closure.  I think I've proven the first part, but am not sure.  My reasoning is as follows:
Consider $X/\bar{B}(x,r) = \{y \in X: d(x,y)>r\}$
$\forall x \in X,\  \exists r>0\ $ s.t. $\ B(x,r)\subset X/\bar{B}(x,r)$
$\therefore X/\bar{B}(x,r)$ is open
$\therefore \bar{B}(x,r)$ is closed
This reasoning seems about as clear to me as the rest of the examples provided in my notes, but to me it seems way too general to work.  Maybe that's my applied/stat background showing, but I'd much prefer more concrete evidence.  Can anyone tell me if I'm headed in the right direction, or am I even more confused than I thought?
Edit: In response to Weaam's answer
Let $\gamma\in X/\bar{B}(x,r)$
Then $d(x,\gamma) > r$
Set $r_{1} = d(x,\gamma) - r$ and consider $B(\gamma,r_{1})$
Let $z \in B(\gamma,r_{1})$
Then $d(x,z)\geq |d(x,\gamma) - d(\gamma,z)|$
$ = |r_{1}+r-d(\gamma,z)| > |r_{1}+r-r_{1}|$
$=|r| = r$
$\therefore z\in X/\bar{B}(x,r)$ and $B(\gamma,r_{1})\subset X/\bar{B}(x,r)$
$\therefore X/\bar{B}(x,r)$ is open
$\therefore \bar{B}(x,r)$ is closed
 A: $\bar{B}(x,r)$ is closed 
Let $y \in X/\bar{B}(x,r)$. Then  $d(x,y) > r$, and we can set $r_1 = d(x,y) - r > 0$ as radius of open ball $B(y, r_1)$. We need to show this radius keeps the open ball inside $X/\bar{B}(x,r)$. If $z \in B(y, r_1)$, then $d(z,x) > r$ (Apply triangular inequality between x,y,z to justify this). Therefore,   $z \in X/\bar{B}(x,r)$
A counterexample to $\bar{B}(x,r) = \overline{B(x,r)}$
The closure $\overline{B(x,r)}$ is the smallest closed set containing $B(x,r)$, i.e. intersection of all closed sets containing $B(x,r)$. So $\overline{B(x,r)} \subset \bar{B}(x,r)$ holds by definition. For the converse, the following exhibits a metric spaces with $B(x,r)$ contained in a closed set stricly smaller than $\bar{B}(x,r)$.
Consider $(X,d)$ with the discrete metric space. Suppose $X \neq \{x\}$, i.e. has more than one element. The open balls in this metric are:
$$B(x,r) = \{y \in X: d(x,y) < r\} = \begin{cases}\{x\} & r \leq 1 \\ X & r > 1\end{cases}$$
Take $r = 1$. Then $\overline{B(x, 1)} = \overline{\{x\}} = \{x\}$ since singletons are closed. 
However, $\bar{B}(x,1) = \{y \in X | d(x,y) \leq 1 \} = X$. Therefore, $\overline{B(x, 1)}  \neq \bar{B}(x,1)$. 
