Closure of $X=\{x:d(x,a)True or False: Can somebody comment on my line of thought?

Closure of $X:=\{x:d(x,a)<r\}$ is always $\bar X:=\{x:d(x,a)\leq r\}$

True:
Suppose for a contradiction it's not. Then we can find a smaller closed set, call it $X_1$ such that $X\in X_1$. Thus we can exclude some point $x_1$ from $\bar X$. Since $X_1$ is closed we can find and open ball: $B_\epsilon (x_1)$ such that $B_\epsilon (x_1) \notin X_1$.
Thus we can find another point $x_2$ such that $x_2 \in B_\epsilon (x_1) \notin X_1$ and also $x_2 \in X$. Thus this is a contradiction, since we assumed $X\in X_1$

Is this correct? I was trying to actually pinpoint $x_2$ by using the
  radius but I couldn't figure it out

 A: Sorry, it's false.
Consider a non empty set $A$ with the discrete metric
$$
d(x,y)=\begin{cases}
0 & \text{if $x=y$}\\
1 & \text{if $x\ne y$}
\end{cases}
$$
Then the closure of $\{x:d(a,x)<1\}$ is $\{a\}$, whereas $\{x:d(a,x)\le 1\}=A$. As soon as $A$ has more than one element, you're doomed.
If your space is $\mathbb{R}^n$ with the Euclidean metric, then it's true.
It is not restrictive to assume $a=0$, which makes things simpler.
The set $\{x:d(0,x)\le r\}$ is obviously closed, so it contains $\bar{X}$. Suppose $d(0,b)=r$ and consider the segment joining $0$ to $b$, which can be seen as the arc $t\mapsto tb$, for $t\in[0,1]$. Now, find a sequence in $X$ converging to $b$.
Hint: $x_n=r(1-1/n)b$
Note that this can be again false if your space is a subset of a Euclidean space with the Euclidean metric. For instance, in the subset $(0,1]\cup\{2\}$, we have
$$
\{x:d(1,x)<1\}=(0,1]
$$
which is (relatively) closed, but
$$
\{x:d(1,x)\le1\}=(0,1]\cup\{2\}
$$
A: Besides the discrete metric, another interesting example is the $p$-adic numbers $\mathbb{Q}_p$, in which the balls $\{x:|x-a|<r\}$ are both open and closed.
