You have found one of the limit points correctly. As a hint for more: What happens if you fix $a$ and let $b\to\infty$?
Remains to show that these are all limit points, i.e. if $x>0$ and $\frac1x\notin \mathbb N$, the $x$ is not a limit point. For such $x$ you can find $n\in\mathbb N$ with $\frac1{n+1}<x<\frac 1n$. Then find $m$ with $\frac1m<x-\frac1{n+1}$. Now if $\epsilon$ is small enough, namely $<\frac1n-x$ and $<x-\frac1{n+1}-\frac1m$, then a great deal of $S$ are seen immediatly to be off by more than $\epsilon$: All with $a\le n$ and $b\le n$ as well was all with $a\ge m$ or $b\ge m$. This should help you see how to make $\epsilon$ a bit smaller if necessary.
To see that $S_1 \cap S_2 = \emptyset$ you only need to remark that an isolated point of $S$ (i.e. a point in $S_2$) is certainly not a limit point of $S$, i.e. a point of $S_1$.
Your other argument goes a long way in proving the equality $\overline{S} = S_1 \cup S_2$, though.
First, $S_1 \subset \overline{S}$, trivially, and $S_2 \subset S$, so certainly $S_2 \subset \overline{S}$, to take care of the inclusion $S_1 \cup S_2 \subset \overline{S}$.
Now (to see $\overline{S} \subset S_1 \cup S_2$): if $x \in \overline{S}$, then it can be a limit point of $S$, and we'd be done, or there is a ball $B(x,r)$ containing only finitely many point of $S$, say $s_1,\ldots, s_n$. If $x$ is not one of the $s_i$, we'd take $r'$ smaller than $r$ and all $d(x, s_i)$ and have a ball $B(x,r')$ around $x$ missing $S$, which cannot be as $x \in \overline{S}$. So $x = s_i$ for some $i$. But now take $r'$ smaller than $r$ and all $d(x, s_j)$, where $j \neq i$, and we have $B(x, r') \cap S = \{x\}$, so $x \in S_2$.
In short, your argument (slightly extended) shows that $x \in \overline{S}$ and $x \notin S_1$, then $x \in S_2$, which shows the required other inclusion.
Best Answer
It seems to me that you have missed the “other than $x$” part of the definition. Clearly, for instance $1$ is not a limit point of, say $\{-1,1\}$.