If a set could be represented as "arbitrary fine" finite union of open balls, then it is not closed If $V$ is a subset of a metric space, such that for every $\varepsilon > 0$ there exists a finite number of open balls $B_{\varepsilon}(x_i)$ such that
$$
 V = \bigcup_{i = 1}^n B_{\varepsilon}(x_i)
$$
then $V$ could not be closed. I conjecture that this is true, but have no idea how to proof this?
 A: This is not true in general. In fact, totally bounded metric spaces are unions of finitely many arbitrarily fine open balls, but are certainly closed in themselves.
Nor is the whole space the only such counterexample. For instance, consider the subset $X=(0,1)\cup(2,3)$ of $\Bbb R.$ Restricting the usual metric of $\Bbb R$ to $X$ makes $X$ a metric space, both $(0,1)$ and $(2,3)$ are unions of finitely many arbitrarily fine open balls, and both are closed in $X$ (though of course, not in $\Bbb R$).
For yet another example, take any infinite set $Y,$ and equip $Y$ with the standard discrete metric $$d(x,y)=\begin{cases}0 & x=y\\1 & x\ne y.\end{cases}$$ Then in fact every subset of $Y$ is open (so every subset of $Y$ is closed), since in particular each singleton $\{x\}$ is the open $d$-ball about $x$ of radius $\varepsilon$ for all $0<\varepsilon<1.$ Every finite subset of $Y$ then provides an example of a closed subset of $Y$ that is a union of finitely many arbitrarily fine open balls.
