If X is a nonempty metric space without isolated points, then X has a dense subset A whose complement is also dense in X Question:

If X is a nonempty metric space without isolated points, then X has a dense subset A whose complement is also dense in X

I can't understand the reason how such $S_{k+1}$ exists in the following note.
http://alpha.math.uga.edu/~pete/Kim99.pdf
Please help
 A: Why $S_{k+1}$ exists is why all the $S_k$ exist:

If $(Y,d)$ is a metric space and $\varepsilon>0$ there exists an $\varepsilon$-net in $Y$, which is a maximal (by inclusion) $\varepsilon$-separated subset $F_{\varepsilon}$ of $Y$, and $A \subseteq Y$ is $\varepsilon$-separated iff $\forall x,y \in A: x \neq y \implies d(x,y) \ge \varepsilon$ 

Proof: let $\mathscr{P}$ be the poset of all $\varepsilon$-separated subsets of $Y$, ordered by inclusion.
Then $\mathscr{P}$ is non-empty, as voidly all singleton sets are $\varepsilon$-separated. It's an inductive poset, because if $\mathscr{A}$ is a chain in $\mathscr{P}$, it's union is still $\varepsilon$-separated (any $x\neq y$ disinct in this union are in a common $A \in \mathscr{A}$ by the chain-property and thus $\ge \varepsilon$ apart.) and an upperbound for $\mathscr{A}$. So Zorn's lemma applies and there is a maximal $M \in \mathscr{P}$.
Note the following property of $M$: 
$$\forall y \in Y: \exists y' \in M: d(y,y') < \varepsilon\tag{1}$$
This holds because if $y \in Y$ and there would be no $y' \in M$ with $d(y,y')< \varepsilon$, then $M \cup \{y\}$ would by definition be $\varepsilon$-separated and strictly larger than $M$, contradicting the maximality of $M$. 
Another noteworthy property is the following: $$\forall y \notin M: \exists \delta_y>0: B(y, \delta_y) \cap M = \emptyset\tag{2}$$ 
which holds as $B(y, \frac{\varepsilon}{2})$ can intersect at most one point of $M$ by the triangle inequality and then taking a smaller $\delta_y$ such that a remaining point is avoided is easily done: note that this implies $M$ is closed in $Y$.
Now the construction in the quoted paper is by recursion. We start with a crowded (no isolated points) metric space $(X,d)$. Let $S_1$ be a $1$-net $F_1$, for $X$, which exists by the previous lemma. 
Now $X\setminus S_1$ can be seen as a metric space in its own right. If $x$ were an isolated point in $X\setminus S_1$, then the fact that $S_1$ is closed (by $(2)$) would imply that $x$ were already isolated in $X$ which cannot be.
So suppose we have constructed $S_0=\emptyset, S_1,\ldots, S_k\subseteq X$ already with the property that 

$\forall k \ge 1: S_k \text{ is an $\frac{1}{k}$-net in } Y_k:=X\setminus \bigcup_{i=0}^{k-1} S_i$ and all $Y_k$ are open in $X$ and thus crowded.

Then $S_{k+1}$ exists in $Y_k$ because of the same argument that $S_1$ exists for $X$, and the same property also holds for $S_{k+1}$: $Y_{k+1}$ is open and is a metric space in its own right and so has an $\varepsilon=\frac{1}{k+1}$-net by the lemma.
The $S_k$ so constructed are clearly pairwise disjoint.
If $A \subseteq \mathbb{N}$ is infinite, then $D:=\bigcup_{n \in A} S_n$ is dense in $X$:
Let $x \in X$ and $\varepsilon >0$. We can suppose WLOG that $x \notin D$ (or we take $y'=x$ in the end). Let $m = \min \{n \in A: \frac{1}{n} < \varepsilon\}$ which exists by the fact that $A$ is infinite. Then $x \in Y_m$ and by property $(1)$ in $Y_m$ we have that there is some $y'\in S_m$, which is an $\frac{1}{m}$-net in $Y_m$, such that $d(x,y') < \frac{1}{m} < \varepsilon$. We have thus found an $y' \in D$ with $d(x,y') < \varepsilon$, which is all we need to show density of $D$ in $X$. 
The arguement is then concluded by noting that taking $A$ the even and odd integers we get two disjoint dense subsets of $X$.
