Check proof: closed ball closure of open ball Consider $\mathbb{R}^n$ with Euclidean metric $d$. Let $\overline{B}(x_0,r)=\{x\in{X}|d(x_0,x)\leq{r}\}$. Show that for any $x\in{\mathbb{R}^n}$ and $r>0$, $\overline{B}(x_0,r)$ is the closure of the open ball ($\overline{B}(x_0,r)=\overline{B(x_0,r)}$). Give an example showing that this is not necessarily true for any metric space.
My proof:
Since $B(x_0,r)$ is open, $\overline{B(x_0,r)}=B(x_0,r)+Bd(B(x_0,r))=\{x\in{X}|d(x_0,x)<r\}+\{x\in{X}|d(x_0,x)=r\}=\overline{B}(x_0,r)$.
My problem is that I didn't use the fact that it's the Euclidean metric, so I can find no example where it doesn't work. Can someone tell me where I went wrong?
 A: 
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 $\{y:d(x,y) \leq 1\}=X \neq \{x\}$

A: We have $\overline{B(x,r)} \subseteq \overline{B}(x,r)$ since $\overline{B(x,r)}$ is the smallest closed set containing $B(x, r)$.
The inclusion $\overline{B}(x,r) \subseteq \overline{B(x,r)}$ holds in every normed space, which $(\mathbb{R}^n, \|\cdot\|_2)$ certainly is:
Take $y \in \overline{B}(x,r)$ and define a sequence $(x_n)_{n=1}^\infty$ as:
$$x_n = \frac{1}{n}\cdot x + \frac{n-1}{n}\cdot y, \quad n\in\mathbb{N}$$
Notice that $(x_n)_{n=1}^\infty$ is a sequence in $B(x, r)$:
$$\left\|x - x_n\right\|_2 = \underbrace{\frac{n-1}{n}}_{< 1}\underbrace{\left\|x - y\right\|_2}_{\le r} < r$$
Also, we have $x_n \xrightarrow{n\to\infty} y$:
$$\left\|y - x_n\right\|_2 = \frac{1}{n}\underbrace{\left\|x - y\right\|_2}_{\le r} \xrightarrow{n\to\infty} 0$$
Thus, $y \in \overline{B(x,r)}$, because it is a limit of a sequence in $B(x,r)$.
In a general metric space, the inclusion $\overline{B}(x,r) \subseteq \overline{B(x,r)}$ need not hold, as the example with the discrete metric shows.
Here is another example on a subset of $\mathbb{R}$. Take the metric space $\big(\langle-1,0]\cup [1, 2], d_2\big)$ where $d_2$ is the standard (Euclidean) metric.
Then $\overline{B}(0,1) = \langle -1,0] \cup \{1\}$, but $\overline{B(0,1)} = \langle -1,0]$. This is because there is no sequence in $B(0,1) = \langle-1,0]$ which would converge to $1$.
