Closure of an $\epsilon-$ball I have the following question:

Is {$x \in X| d(x,y) \le \epsilon $ } closed in $X$ for every metric space  $(X,d)$, every $y \in X$ and every $\epsilon > 0$?

In my opinion it is, because {$x \in X| d(x,y) \le \epsilon $ } is basically just the closure of an $\epsilon-$ball. I just wanted to check, if my answer is correct, because I heard some other students saying they found a counterexample to disprove it. If I'm wrong, could someone tell me why?
Thanks in advance.
 A: Yes, $\{x\in X:d(x,y)\leq \varepsilon\}$ is closed for all $y\in X$ and all $\varepsilon>0$. One way to prove this is to show that the function $x\mapsto d(x,y)$ is continuous on $X$ for each fixed $y\in X$.
However, it is not always the case that $\{x\in X:d(x,y)\leq \varepsilon\}$ is the closure of $\{x\in X:d(x,y)< \varepsilon\}$. For instance, if $d$ is the discrete metric then $\{x\in X:d(x,y)<1\}$ is closed and is strictly contained in $\{x\in X:d(x,y)\leq 1\}$
A: Let $$ y\not \in \bar{B}(x,\varepsilon)=\{x \in X| d(x,y) \le \epsilon\}~~ i.e ~~~d(x,y)>\varepsilon .$$
for $$r=\frac{d(x,y)-\varepsilon }{2}>0$$  we have 
$$B(y,r)\subset \bar{B}(x,\varepsilon)^c \implies\bar{B}(x,\varepsilon)^c~~is ~open~~\Longleftrightarrow \bar{B}(x,\varepsilon) ~~~is ~~closed$$
Indeed, for $z\in B(y,r)$ since  $d(x,y)>\varepsilon $ we have 
$$d(x,y)\le d(x,z)+d(y,z) <d(x,z) +r\\\implies d(x,z)> d(x,y)-r =d(x,y) -\frac{d(x,y)-\varepsilon }{2}=\frac{\varepsilon }{2} +\frac{d(x,y)}{2} >\varepsilon $$
that is $$ d(x,z)>\varepsilon \implies z\not\in \bar{B}(x,\varepsilon)^c$$
Therefore,
$$B(y,r)\subset \bar{B}(x,\varepsilon)^c \implies\bar{B}(x,\varepsilon)^c~~is ~open~~\Longleftrightarrow \bar{B}(x,\varepsilon) ~~~is ~~closed$$
