TITLE: $(X,d)$ is separable if and only if $X$ does not contain an uncountable closed discrete subset
QUESTION [8 upvotes]: The problem is to prove that a metric space $(X,d)$ is separable if and only if it does not contain an uncountable closed discrete subset.
The first thing I want to note is that being closed and discrete is equivalent to not having any limit points, so the problem can be restated as "prove $(X,d)$ is separable if and only if every uncountable subset has a limit point in $X$".
The forward direction is simple enough. Suppose $X$ is separable and let $A$ be a countable dense subset. Assume to the contrary that there does exist an uncountable subset $B$ which has no limit points. Then for each $b\in B$, there exists $\epsilon_b>0$ such that the open ball $B_d(b,\epsilon_b)$ is disjoint from $B\setminus\{b\}$. Then the collection of open balls $\mathcal{A}=\{B_d(b,\epsilon_b/2):b\in B\}$ is an uncountable collection of pairwise disjoint open sets. Since $A$ is countable and $\mathcal{A}$ is not, $A$ cannot intersect every open ball in $\mathcal{A}$, so there must exist $b\in B$ such that $B_d(b,\epsilon_b/2)$ is disjoint from $A$ contradicting the fact that $A$ is dense in $X$.
It's the other direction that's giving me trouble. Suppose that every uncountable subset of $X$ has a limit point in $X$. I need to construct a countable dense subset (or use a nonconstructive existence proof method such as Zorn's Lemma). I have tried coming up with ways of constructing such a countable dense subset, I have tried proving the contrapositive (which is also an existence proof), I have tried using Zorn's Lemma, but nothing I have tried worked. Does anyone have hints or ideas?

REPLY [2 votes]: Okay, so I think I have a solution to my question. Here is an overly detailed proof:
Suppose $X$ does not contain an uncountable subset with no limit points so that every subset with no limit points is countable. Let $n\in\mathbb{Z}^+$ and let $\mathcal{A}_n=\{Y\subset X:\forall x,y\in Y, d(x,y)\geq\frac{1}{n}\}$. I claim $\mathcal{A}_n$ has a maximal element under the subset relation. Let $\{C_i:i\in I\}$ be a totally ordered chain in $\mathcal{A}_n$ and consider $\cup_{i\in I}C_i$. Let $x,y\in \cup_{i\in I}C_i$ so that $x\in C_i$ and $y\in C_j$ for some $i,j\in I$. Since $\{C_i:i\in I\}$ is totally ordered, either $C_i\subset C_j$ or $C_j\subset C_i$. Assume without loss of generality that $C_i\subset C_j$ so that $x,y\in C_j$. Since $C_j\in\mathcal{A}_n$, this means $d(x,y)\geq\frac{1}{n}$. Thus $\cup_{i\in I}C_i\in\mathcal{A}_n$ and is an upper bound on $\{C_i:i\in I\}$ under the subset relation, so by Zorn's Lemma, $\mathcal{A}_n$ has a maximal element. Let $D_n$ denote such a maximal element for each $n\in\mathbb{Z}^+$.
Now, I claim $D_n$ has no limit points. Assume $D_n$ has a limit point $x\in X$. Then the open ball $B_d(x,\frac{1}{2n})$ must intersect $D_n\setminus\{x\}$. Note that $B_d(x,\frac{1}{2n})$ cannot contain more than one element of $D_n$, for any two elements in $B_d(y,\frac{1}{2n})$ must be within a distance $\frac{1}{n}$ of each other by the triangle inequality. Let $y$ be the single element of $B_d(x,\frac{1}{2n})\cap D_n\setminus\{x\}$. Then $B_d(x,d(x,y))$ is an open neighborhood of $x$ which is disjoint from $D_n$ contradicting the assumption that $x$ is a limit point of $D_n$. Thus, $D_n$ contain no limits points which, by assumption, means that $D_n$ is countable.
Finally, I claim $D=\cup_{n\in\mathbb{Z}^+}D_n$ is a countable dense subset of $X$. We see immediately that $D$ is countable since it is a countable union of countable sets. Let $x\in X$ and assume $x\notin\overline{D}$. Then there exists $\epsilon>0$ such that $B_d(x,\epsilon)$ is disjoint from $D$. Let $n\in\mathbb{Z}^+$ such that $\frac{1}{n}\leq\epsilon$. Then, because $B_d(x,\epsilon)\cap D=\emptyset$, for all $y\in D_n$, $d(x,y)\geq\frac{1}{n}$. Then that implies $D_n\cup\{x\}\in\mathcal{A}_n$, contradicting the maximality of $D_n$. Thus, we conclude that $D$ is a countable dense subset of $X$ so that $X$ is separable.