Can an open ball be closed if the open ball contains infinite points? Consider a metric space $(X,d)$. Is the following statement true?
An non-finite open ball in X with finite radius is never closed. Non-finite in this sense means that the open ball contains an infinite amount of points.
 A: $B(0,1)=(-1,1)$ in the space $\mathbb{R} \setminus \{ -1,1 \}$ with the standard metric is a counterexample. $B(x,2)$ for any $x$ in an infinite discrete metric space is another counterexample. 
These two show the two possible ways that this can happen: either the space is disconnected (in which case the ball can be a proper subset of the space) or the space is bounded (in which case the entire space can be written as an open ball). So this is impossible in an unbounded connected space, which includes among other things all vector spaces over $\mathbb{R}$ or $\mathbb{C}$ with positive dimension.
A: No, this is not true. In $\mathbf R$ with the discrete metric, the open ball $U_2(0)$ of radius $2$ is the whole space $\mathbf R$, hence open-closed and infinite.
If you want a non-discrete example, consider $\mathbf R$ with the metric 
$$ d(x,y) = \left| \arctan x - \arctan y \right| $$
and $U_{7}(0) = \mathbf R$.
A: Counterexample: Consider the space $X = [0,2] \cup [4,5]$ (with the usual $|\cdot|$ metric).
Consider the balls of radius $2$ centered at $1$.
A: Consider the unit circle as embedded in $\mathbb{R}^2$. Any ball therein with radius $>\pi$ is the entire space and hence closed.
A: Let $X=\Bbb Q$, and let $d$ be the usual metric. For every $p\in\Bbb Q$ and every irrational $r>0$, the ball $B(p,r)=\{q\in\Bbb Q:|q-p|<r\}$ is both open and closed in $\Bbb Q$.
A: When $d$ is a metric then $e_1=d/(1+d)$ and $e_2=\max (d,1)$ are metrics equivalent to $d,$ that is, they generate the same topology that $d$ does. Since $\sup e_1\leq 1\geq \sup e_2,$ an open $e_1$-ball or   $e_2$-ball of radius $2$ is the whole space, which is both open and closed, regardless of whether it is a finite or infinite space.
Another common misconception is that $\{y:d(y,x)\leq r\}$ is always the closure of $\{y:d(y,x)<r\}.$
