The function $f: (\mathbb{R}^n,d) \to (\mathbb{R},|.|)$ where$$f(x_1,x_2...x_n)=\sqrt{(x_1-x_{0,1})^2+...+(x_n-x_{0,n})^2}$$ is continuous thus the set $\{x:d(x,x_0) \leq r\}$ is closed as the inverse image of of the set $[0,r]$ under a contunuous function.
If you are not familiar with the topological definition and properties of continuity then ignore all the previous and just prove that the set $\{x:d(x_0,x) \leq r\}$ is closed using the sequential characterization of closedness.
Now $B(x_0,r) \subseteq \{x:d(x,x_0) \leq r\}$ thus $\overline{B(x_0,r)} \subseteq\{x:d(x,x_0) \leq r\}$ because the closure is the intersection of all closed sets that contain $B(x_0,r)$ as a subset.
Now let $z=(z_1,z_2...z_n) \in \{x:d(x,x_0) \leq r\}$
$\bullet$ If $d(z,x_0)<r$ then $z \in B(x_0,r) \subseteq \overline{B(x_0,r)}$
$\bullet$ If $d(z,x_0)=r$ then consider the sequence $z_m=(\frac{m}{m+1})z+\frac{1}{m+1}x_0$.
We have that $$\sqrt{(\frac{m}{m+1}z_1+\frac{x_{0,1}}{m+1}-x_{0,1})^2+....+(\frac{m}{m+1}z_n+\frac{x_{0,n}}{m+1}-x_{0,n}})^2$$ $$=\sqrt{(\frac{m}{m+1})^2(z_1-x_{0,1})^2+...+(z_n-x_{0,n})^2)}$$ $$=\frac{m}{m+1}d(z,x_0)=\frac{m}{m+1}r<r$$ thus $z_m \in B(x_0,r), \forall m \in \mathbb{N}$
Also we have that $z_n \to z$ with respect to $d$
So for every $z \in \{x:d(x_0,x) \leq r\}$ such that $d(x_0,z)=r$ we found a sequece $z_m \in B(x_0,r)$ such that $z_m \to z$
Thus $z \in \overline{B(x,r)}$
From the two bullets we conclude that $\{x:d(x_0,z) \leq r\} \subseteq \overline{B(x_0,r)}$
Now as a counterexample in a general metric space:
Take a discrete metric space $X$ with more than one elements.
Let $x \in X$
Then $\overline{B(x,1)}=\overline{\{x\}}=\{x\}$
but $\{y:d(x,y) \leq 1\}=X \neq \{x\}$
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
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.