The statement of the problem is not so correct, since we are allowed to choose $p,q \in S$ such that $p=q$, viz. $d(p,q)=0$, so from here later we'll assume $p\neq q$. Let's go back to the problem now: since we assume the contraint $d(p,q)>r/2$ then $S$ is finite, i.e. there exists a integer $n\ge 1$ such that $|S|=n$ and $S:=\{p_1,p_2,\ldots,p_n\}$. It's well known that the propositions:
i) $X$ is totally limited
ii) $X$ is compact
iii) $X$ is sequentially compact
are equivalent (the easiest way to prove it is that (i) implies (ii) implies (iii) implies (i)).
So, there exists a finite collection of open balls $B(x_1,3r/4),B(x_2,3r/4),\ldots,B(x_{k_1},3r/4)$ that covers the whole compact metric space $(X,d)$. We can also assume without loss of generality that $B(x_i,3r/4) \cap X \neq \emptyset$ for all $1\le i\le k_1$.
Define $x_1:=\alpha_1$, and also $X_1:=X\setminus B(x_1,3r/4)$: if $x_1$ is empty then we ended (see below), otherwise $X_1\neq \emptyset$ is a metric space too, bounded, and closed, hence compact too. Then there exist a finite collection of open balls $B(y_1,3r/4),B(y_2,3r/4),\ldots,B(y_{k_2},3r/4)$ that covers $X_1$. Define $y_1:=\alpha_2$.
Repeat this algorithm infinitely many times, we have two cases:
1) If the sequence $\alpha_1,\alpha_2,\ldots$ is finite, then just define $p_i:=\alpha_i$ for all $i$ and we are done, indeed $d(p_i,p_j)\ge 3r/4 > r/2$ for all $1\le i < j \le n$.
2) If the sequence $\alpha_1,\alpha_2,\ldots$ is not finite, then the infinite collection of open balls $B(\alpha_1,3r/4),B(\alpha_2,3r/4),\ldots$ is a cover of $X$. Since $X$ is compact there exists a finite set of pairwise disjoint positive integers $T:=\{t_1,t_2,\ldots,t_n\}$ such that $B(\alpha_{t_1},3r/4),B(\alpha_{t_2},3r/4),\ldots,B(\alpha_{t_n},3r/4)$ is a cover too. Just set $\alpha_{t_i}=p_i$ for all $1\le i\le n$ and we really made our subcover of open balls that "do not overlap too much". []
In a metric space, every open ball is an open set, but certainly not the other way around. A trivial example is simply to take the union of two disjoint open balls (in, say $\mathbb R^2$ if you want to get a nice picture). Most certainly the union of two disjoint open balls is not an open ball, but it is an open set.
To solve your problem, think of closure properties of $\sigma$-algebras, of open sets and open balls (in separated metric spaces), and the relationship between open/close and set complementation.
Best Answer
$B(a,r)$ is just the solution of $\left|\dfrac1a -\dfrac1x\right|<r$, which is easily solved:
$ x>\dfrac{a}{1+a r} \qquad \qquad \qquad\text{if } r\ge\dfrac1a $
$ \dfrac{a}{1+a r}<x<\dfrac{a}{1-a r} \qquad \text{if } r<\dfrac1a $
In the first case, the ball is infinite. In the second case, the ball is finite.
In particular, the ball is infinite if $r\ge1$ and the ball centered at $1$ is all of $\mathbb N$ if $r=1$.