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". []
For $Y\subseteq X$, this means that the subset $Y$ is a compact space when considered as a space with the subspace topology coming down from $X$.
To jog your memeory, recall that the subspace topology works this way: the open sets of $Y$ are just the intersections of $Y$ with open sets of $X$.
This turns out to be equivalent to the following condition: given any open covering of $Y$ by open sets of $X$, there is a finite subcollection covering $Y$.
I think you'll be able to prove that equivalence rather easily. I am also unsure of how the terminology of "covers" jives with what you learned. I can clear up whatever questions you have about what I meant in the comments. Just ask :)
I hope this takes care of your question about definitions. If you need further help with the question you're working on, I can suggest another problem/answer posted at this site: Does a compact subspace have to be closed in an arbitrary metric space?.
Best Answer
Right: you cannot prove this starting from a specific cover.
If $\mathcal U$ is an open cover of $K$, there is some $A_0\in\mathcal U$ such that $0\in A_0$. Since $A_0$ is open and $\lim_{n\to\infty}\frac1n=0$, $\frac1n\in A_0$ is $n$ is large enough. So, there is some $N\in\mathbb N$ such that $n\geqslant N\implies\frac1n\in A_0$. For each $n\in\{1,2,\ldots,N-1\}$, let $A_n\in\mathcal U$ be such that $\frac1n\in A_n$. Then $\bigl\{A_n\,|\,n\in\{0,1,\ldots,N-1\}\bigr\}$ is a finite subcover of $\mathcal U$.