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.
Let $\mathbb{R}$ be given the usual metric.
The idea of limit points of a subset $A \subset \mathbb{R}$ is to say that in the neighbourhood of such points, I can find at least one other point different from that point which is also in $A$.
So if $x \notin A$, then $x$ is a limit point of $A$ iff in any $\varepsilon$-neighbourhood of $x$, I can find at least one point of $A$. This is precisely your modified definition. So yes, it should be valid.
Addendum:
For the edited part: Note the condition $a_n \neq x$ for all $n \in \mathbb{N}$ carefully. This condition precisely prevents the constant sequence that you mentioned.
If this condition is not present, then every point in a set $A$ could be a limit point of $A$. This is not necessarily true. Consider the following set
\begin{equation}
S = \{ 1/n : n \in \mathbb{N}\}.
\end{equation}
Then none of the points in $S$ is a limit point of $S$. Moreover, the only limit point of $S$ is $0$ which is not in $S$.
Finally, the definition does not fail with the constant sequence as it is sufficient to find just one such sequence that satisfies $x = \lim a_n$.
Best Answer
As noted above, the argument you have in mind works in the case of the union of two sets. Consequently, you can infer that this property holds for finite unions. However, it does not hold in the case of infinite unions. For example, in the standard topology on $\mathbb{R}$, $\bigcup_{n \in \mathbb{N}} [1/n, 1] = (0, 1]$ while the closure of this union is $[0, 1]$.