Closure of a Ball is Contained in Closed Ball (usual metric) 
Let $\overline{B_d (x, r)}$ be the closure of an open ball, and let $\hat B_d (x, r)$ be a closed ball. Show that in any metric space $(X, d)$, $\overline{B_d (x, r)} \subseteq \hat B_d (x, r)$ for all $x \in X$ and $r > 0$. 

I used the definition of closure: $\overline{B}$ = $B \bigcup B'$. If $x \in \overline{B}$, then $x \in B$ or $x \in \overline{B}$, or both. If $x \in B$, then clearly $x \in \hat B$ since $B \subseteq \hat B$. But, if $x \in B'$, then $x$ is an accumulation point. Since $\hat B$ is closed, it contains all accumulation points of $B$. Thus, $x \in \hat B$.
Is this approach correct, or at least in the right direction? Should I be using the closure definition, or try another one? Thank you.
 A: First note that in any metric space $(X,d)$ the closed ball 
$D(x,r) = \{y \in X:: d(x,y) \le r\}$ is a closed set. (Hence the justification of the name "closed ball")
To see this let $y \notin D(x,r)$. We will find an open ball around $y$ that is disjoint from $D(x,r)$ and this will show (as we can do this for all such $y$) that $D(x,r)$ is closed.
We have from $y \notin D(x,r)$ that $d(y,x) > r$, so define $s = d(y,x) - r > 0$.
Claim: $B(y,s) \cap D(x,r) = \emptyset$, fulfilling the promise. 
Suppose that there exists some $z \in B(y,s) \cap D(x,r)$. Then $d(y,z) < s$ and $d(x,z) \le r$ by the sets' definitions. But then by the triangle inequality:
$$d(y,x) \le d(y,z) + d(z,x) < s + d(z,x) \le s + r =d(y,x)$$ which is a contradiction as a number can never be strictly smaller than itself.
So no such $z$ exists and the intersection is empty. 
As $B(x,r) \subseteq D(x,r)$ and the latter set is closed we always have $\overline{B(x,r)} \subseteq D(x,r)$. (As the closure is the smallest closed subset containing a set.) 
