If $y_n$ is not convergent and unbounded, $y_n \to \infty$. But for $k > \frac{4}{x}$, note that $X_{k}$ lies between $[0, \frac{2}{k})$, while for $\epsilon<\frac{x}{2}$, $\epsilon$ neighbourhoods lie in $(\frac{x}{2}, \infty)$. Thus a contradiction, as $Y_k$ is empty if $k > \frac{4}{x}$. We repeat a similar logic for $x<0$. Thus, every limit points of $X$ are of the form $\{\frac{1}{n} \mid n \in \mathbb{Z} \setminus {0} \} \cup \{0\}$
$Y_n$ is not empty for any $n$ since $x$ is positive and is a limit point. By the well-ordering property, the sequence $y_n=\min Y_n$ is well defined. Note that $Y_{n+1} \subset Y_n$ so we have $y_n \le y_{n+1}$. If $y_n$ is bounded, it thus converges to a natural number $y$ as all $y_n$ are natural numbers. We may say that $x$ is a limit point of $X_y$, and by Step 1 we have that $x=\frac{1}{y}$.
Then, I supposed $x$ was a limit point of the given set and let $Y_n$ denote set of natural $m$'s for which $X_m$ and $B_{\frac{1}{n}}(x) \setminus \{x\}$ have non-empty intersections.
Yeah, so the argument I used was first that $X_n=\{ \frac{1}{n} +\frac{1}{m}\ \mid |m|\ge |n|, m \in \mathbb{Z}\}$ for fixed, non-zero $n$ has only $\frac{1}{n}$ as it's limit point.
