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Well, it's a known fact that the closure of an open ball is not necessarily the closed ball with the same center and radius, but I assume that then it follows that the closure is necessarily a subset of the closed ball with the same center and radius. I have seen lots of counterexamples on this site but I am seeking for one that works in the entirety of $R_2$ with a newly defined distance, not an example that eliminates some parts of $R_2$.

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    What's $R_2$? Did you mean $\mathbb{R}^2$? – 2017-01-29
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    Yes sorry. For istance, a typical counterexample is to consider an open ball of radius 1 with the discrete distance. The closed ball is the entire set, while the closure would be only the point in which the open ball is centered itself, but I don't get why the closure of the open ball is also the only point itself as the open ball... – 2017-01-29
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    Since every singleton set is an open ball, it follows that the complement of a singleton set, being a union of singleton sets, is a union of open sets, hence is open. Since the complement of a singleton set is open, a singleton set is closed (though it's also open). So for any $p$ in $\mathbb{R}^2$, $B(p,1)=\{p\}$, but $\{p\}$ is closed, hence the closure of $B(p,1)$ is $\{p\}$, However $B[p,1]$ is all of $\mathbb{R}^2$. – 2017-01-29

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Consider $\Bbb R^2$ with the discrete metric. We have $B(0,1) = \{0\}$, so $\overline{B(0,1)} = \{0\}$. However, $B[0,1] = \Bbb R^2$.

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    Your username fits this question perfectly. – 2017-01-29
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    I follow you when you say what are the open ball and the closed ball, but how did you find that the closure of the open ball is also the point itself? – 2017-01-29
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    @John because in any metric space, singletons are closed. Indeed, if $a \in \{x\}^C$, i.e. $a\notin \{x\}$, then $d(x,a) = r > 0$, so $B(a,r) \subset \{x\}^C$. Hence $\{x\}^C$ is open, i.e. $\{x\}$ is closed. – 2017-01-29