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". []
Let $X$ be an infinite discrete space with the usual discrete metric, $d(x,y)=1-\delta_{xy}$. Since $X$ is infinite and discrete $\{x\}_{x\in X}$ is an open cover with no finite subcover. Cauchy sequences in $X$ are eventually constant, and thus converge, and any sufficiently small closed ball contains a single point and is thus compact. Note that $X$ is the closed ball of radius 1 around any point. Hence $X$ satisfies your conditions.
Edit: Some comments on my thought process, in response to the question in the comments.
My first thought was that we know that a metric space is compact if and only if it is complete and totally bounded. Since our whole metric space is complete here, we need to find a way to make small closed balls totally bounded and large balls not totally bounded. The way I think about total boundedness, this is similar to saying we want small neighborhoods of a point to be "finite dimensional" and large neighborhoods to be "infinite dimensional." In a very rough intuitive way.
From there, the discrete topology is fairly natural, since small balls are literally finite and large balls contain the whole space.
As for my thoughts on your thoughts, well, they're correct, but I'm not sure where you would go from where you got to.
Best Answer
In a compact metric space the general answer is no.
In fact, your condition has a name. A metric space $X$ such that every ball of radius $r$ can be covered by $k$-many balls of radius $\frac{r}{2}$ is called doubling. It turns out the doubling condition is equivalent to having finite Assouad dimension, which is stronger than having finite Hausdorff dimension, which in turn is stronger than having finite topological dimension.
Also any space with a doubling measure $(\mu(2B)\leq C\mu(B)$ for some constant $C$, for every ball $B$) must have a doubling metric, which is an easy exercise.
Therefore any metrizable compact abelian group with infinite topological dimension will do as a counter-example. For example, the countably infinite product of the circle with itself $\left(\mathbb R/\mathbb Z\right)^{\mathbb N}$ cannot have a doubling metric, thus its Haar measure cannot be doubling with respect to an invariant metric (or any compatible metric, for that matter).