Separable spaces & fixed distance subsets How can I prove the following?
Claim: let $X$ be a metric space containing a dense set $S$. Assume that any subset $F_K$ of $X$ such that the elements of $F_K$ are at least at distance $K>0$ from each other pairwise is countable. Then $X$ is separable.
Intuition: If there exists an uncountable subset such that the elements are at distance at least a fixed $K$ from each other, then $X$ is not separable. The contrapositive is: $X$ is separable implies that for all sets such that the elements are at least at some fixed distance from each other, these sets are countable. I am trying to prove the converse.
Proof: Not sure how to proceed. Here’s a failed attempt since the $F_{K_n}$ are not well-defined. Consider the sequence $K_n=2^{-n}$. For each $K_n$, fix an enumeration $x^n_j$ of the elements in the countable set $F_{K_n}$. Then, find an approximating sequence $(f^{j,n}_m)_{m\ge1}$ of each element $x^n_j$  in $F_{K_n}$. We then claim that the set $S’=\{f^{j,n}_m | j,n,m\ge1\}$ is a countable set which is dense in $X$. It is obviously countable. To show that it is dense, pick an $x \in X$. Then, for each $\varepsilon =  1, 2^{-1}, 2^{-2}, \ldots$ we may pick another element… [here, I realised that my proof was wrong].
 A: Your idea is good, but I don't understand the proof.
Try this:
Since $F_{1/n}$ is countable for all $n\in\mathbb{N}$, then $D=\bigcup_{n\in\mathbb{N}}F_{1/n}$ is countable too. Next we can prove that $D$ is dense in $X$, indeed, given $x\in X$ there are two possibilities: $x\in D$ or $x\notin D$.
If $x\in D$, then trivially we have the constant sequence $x_n=x$ in $D$ such that $x_n\to x$.
If $x\notin D$, then $x\notin F_{1/n}$ for all $n\in\mathbb{N}$, and hence, for each $n\in\mathbb{N}$ there exists $x_n\in F_{1/n}$ such that $d(x,x_n)<1/n$, but this means that there exists a sequence $(x_n)\subset D$ such that $x_n\to x$.
A: Suppose all fixed-distance sets are at most countable in $(X,d)$. We'll show that $X$ is separable.
For each $n$ by your favourite maximal principle (e.g. Zorn's lemma) pick a maximal (by inclusion) family $D_n \subseteq X$ that is an $\frac1n$-distance set. So if $x \notin D$ we know there is some $y \in D_n$ such that $d(x,y) < \frac1n$ (or $D_n \cup \{x\}$ would contradict the maximality of $D_n$) and for $x \in D_n$ we can just pick $y=x$ to get this inequality. So in short:
$$\forall x \in X:\exists y(x) \in D_n: d(x,y(x)) < \frac1n\tag{1}$$.
It follows from $(1)$ that $D:=\bigcup_n D_n$ is dense in $(X,d)$ (for any open ball $B(x,r)$ we find $y(x) \in D_m$ for some $m$ with $\frac1m < r$ and so all open balls intersect $D$ etc.) and $D$ is countable as a countable union of at most countable sets (by the assumption on distance set sizes).
QED.
IMHO this is much simpler than your attempt.
