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.
A set $C$ in a topological space $X$ is closed if for every $x\in X\setminus C$, there exists an open set $U$ such that $x\in U \subseteq X\setminus C$.
A point $x$ in a subspace $S$ of a topological space $X$ is an isolated point of $S$ if there exists an open set $U$ such that $U\cap S = \{x\}$.
A point $x$ is a limit point of a subspace $S$ of a topological space $X$ if every open set $U$ which contains $x$ satisfies $U\cap (S\setminus\{x\})\ne\emptyset$.
Let's notice something right away:
Let $X$ be a topological space, $S\subseteq X$, and $x\in S$. Then $x$ is an isolated point of $S$ iff $x$ is not a limit point of $S$.
Proof:
Since $x\in S$ by hypothesis, we have that $x$ is an isolated point of $S$ iff there exists $U$ open such that $U\cap S=\{x\}$ iff there exists $U$ open containing $x$ such that $U\cap(S\setminus\{x\})=\emptyset$ iff $x$ is not a limit point of $S$. $\square$
So if we have a closed ball $B$ and a point $x$ in $\mathbb{R}^n\setminus B$, then the union $A:=B\cup\{x\}$ is a closed set, $x$ is an isolated point of $B$, and $x$ is not a limit point of $B$.
Best Answer
It's not closed in general: take the set $\{\frac{1}{n} : n \in \mathbb{N} \}$.