The standard counterexample is to use the discrete metric on the space $X$: $||x - y|| = \begin{cases}0 ,& x = y \\ 1 ,& x \neq y \end{cases}$. The open unit ball around $x$ is $\{x\}$. The closed unit ball around $x$ is $X$. Your theorem fails when you claim "$x \in \overline{B(x_0,r)}$" because you have assumed that $B(x_0,r)$ actually contains points, $x$, such that $0 < ||x - x_0|| < 1$, which is false in this counterexample.
Let $D(x,r) = \{ y | \|x-y\| \le r \}$ and $B(x,r) = \{ y | \|x-y\| < r \}$.
We have $B(x,r) \subset D(x,r)$ for all $r$.
Since the norm is continuous we see that $D(x,r)$ is closed and hence $\overline{B}(x,r) \subset D(x,r)$.
Now suppose $y \notin \overline{B}(x,r)$. Since $\overline{B}(x,r)$ is closed
there is some $\epsilon>0$ such that $B(y,\epsilon)$ does not intersect $\overline{B}(x,r)$ (and hence does not intersect $B(x,r)$). It follows that
$\|x-y\| \ge r + \epsilon$ and
so $y \notin D(x,r)$. Hence $D(x,r) \subset \overline{B}(x,r)$.
Addendum: To show why $\|x-y\| \ge r + \epsilon$:
Let $\phi(t) = t y +(1-t)x$. For $0 \le t_1 \le t_2 \le 1$ we have $\|\phi(t_2)-\phi(t_1)\| = (t_2-t_1) \|x-y\|$.
Note that $\phi(t) \in B(x,r)$ if $t \in [0,{r \over \|x-y\|})$ and
$\phi(t) \in B(y,\epsilon)$ if $t \in (1-{\epsilon \over \|x-y\|}), 1]$, and we must have $t_1={r \over \|x-y\|} \le t_2=1-{\epsilon \over \|x-y\|}$ since the balls do not overlap.
Finally $\|x-y\|= \|\phi(0)-\phi(1)\| \ge \|\phi(0)-\phi(t_1)\| + \|\phi(t_2)-\phi(1)\| = r+ \epsilon$.
Another take:
Consider $\phi(t)$ for $t \in [0,1]$. For the $t \in [0,{r \over \|x-y\|})$ part we have $\phi(t) \in B(x,r)$ and for $t \in (1-{\epsilon \over \|x-y\|}), 1]$ we have $\phi(t) \in B(y,\epsilon)$. The balls do not overlap, so the intervals $[0,{r \over \|x-y\|}), (1-{\epsilon \over \|x-y\|}), 1]$ are disjoint and the combined length of the intervals is ${r+ \epsilon \over \|x-y\|}$.
Since $\|\phi(t_2)-\phi(t_1)\| = (t_2-t_1) \|x-y\|$ for $0 \le t_1\le t_2 \le 1$ we see that
$\|x-y\|=\|\phi(1)-\phi(0)\| = (1-t_2)+(t_2-t_1)+(t_1-0)) \|x-y\| \ge (1-t_2)+(t_1-0) \|x-y\|$, and since $(1-t_2)+(t_1-0) = {r+ \epsilon \over \|x-y\|}$ we have the desired result.
Best Answer
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.)