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.
Suppose $A \subseteq X$ and $x$ is a limit point of $A$, and $X$ is Hausdorff.
This means that for every open neighbourhood $U$ of $x$, $U$ intersects $A \setminus \{x\}$.
Suppose now that $U \cap A$ were finite for some open $U$.
Fact: in a Hausdorff space all finite sets are closed. (In fact the latter is equivalent to $X$ being $T_1$ and Hausdorff implies $T_1$) so $V:= X\setminus ((U \cap A) \cup \{x\})$ is open and contains $x$. Now $(V \cap U) \cap A \subseteq \{x\}$ : $y \in (V \cap U) \cap A$ implies $y \in U \cap A$ but as $y \in V$ too, $y=x$. This shows that the open neighbourhood $U \cap V$ of $x$ witnesses that $x$ is not a limit point of $A$, contrary to assumption.
So $U \cap A$ can never be finite, for $U$ an open neighbourhood of $x$.
We only use Hausdorff mildly; $T_1$ is all we need.
Best Answer
A point in a closed set is either a limit point or an isolated point (a point which has a neighbourhood which contains no other points of the set). For a reference, this is Theorem 17-E (page 97) of Introduction to Topology and Modern Analysis by G. F. Simmons.