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". []
If no finite subcollection of $\{G_{n}\}$ covers X, then $F_{1} = G_{1}^{c}\neq\emptyset$, so choose $x_{1}\in F_{1}$. Having chosen $x_{1}, . . .,x_{j} \in X$, choose $x_{j+1} \in X$, if possible, so that $x_{j+1} \in F_{j+1} \setminus \{x_{1}, ...,x_{j}\}$. If I prove that this process must not stop after a finite number of steps, then you will get distinct $x_{n}'s$ and the set E will be infinite.
suppose this process stops after a finite number of steps, say k, then $F_{k+1}\setminus \{x_{1},...,x_{k}\}=\emptyset$. Then $G_{1}\cup ....\cup G_{k+1}\cup\{x_{1},...,x_{k}\}=X$. Since $\{G_{n}\}$ covers X, there exists $G_{m_{i}}$ such that $x_{i}\in G_{m_{i}}$ for $i = 1,...,k$. Now $G_{1}\cup ...\cup G_{k+1}\cup G_{m_{1}}\cup ...\cup G_{m{k}} = X$, a contradiction.
Best Answer
First a notational observation: the notation $\{x\}^{(n)}$ is truly horrible. It looks as if you’re performing some operation on the singleton set $\{x\}$ whose sole element is $x$.
Your basic idea is just fine, but you’re making it too complicated; that’s part of why you’re having trouble expressing it clearly. As others have suggested, you don’t need to try to relate the centres of one ‘level’ to those of any other ‘level’. Here’s a relatively efficient version of the idea:
You might like to note, by the way, that we could have got the same result had each of the sets $F_n$ been countable: we did not actually need them to be finite. Thus, the same argument shows that every Lindelöf metric space is separable. And this actually is a stronger result, since $\Bbb R$ with its usual metric is Lindelöf but not compact.