A closed set in $Z$ is finite (or the whole set $Z$). Thus the only closed set containing $E$ is $Z$ and therefore the closure of $E$ is $Z$: the closure of a set is the intersection of all closed sets containing it.
The interior of a set is the largest open set contained in it. So, let $U$ be an open set, $U\subseteq E$; then $Z-U$ is closed, but it's clearly infinite, because $Z-U\supseteq Z-E$. The only infinite closed set in $Z$ is $Z$; therefore $Z-U=Z$ and $U=\emptyset$.
Similarly, $A$ contains no nonempty open set and therefore the interior of $A$ is again empty. Of course $A$ is closed, by definition of cofinite topology.
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
To show that $E'$ is closed, we will show $X\setminus E'$ is open. If $x \in X\setminus E'$, then there is $B(x,r) \subseteq X\setminus E$. Then for all $y \in B(x,r)$ since $B(x,r)$ is open there is $B(y,r_2)\subseteq B(x,r)\subseteq X \setminus E$ so $y \in X \setminus E'$ as well. It follows that $x$ is an interior point of $X\setminus E'$, and so $X\setminus E'$ is open, and $E'$ is closed.
Now we show $E$ and $\overline{E}$ have the same limit points. Clearly $E' \subseteq \overline{E}'$ since if every $B(x,r)$ contains an element of $E$ different from $x$, then that element is also an element of $\overline{E}$. Conversely, suppose $B(x,r)$ contains an element $y$ of $\overline{E}$. Since $B(x,r)$ is open, we may choose $B(y,r_2)\subseteq B(x,r)$ and by making $r_2<d(x,y)$ smaller if necessary we may ensure $x\notin B(y,r_2)$. Then since $y \in \overline{E}$ we may choose $z \in E \cap (B(y,r_2)\setminus\{y\})$. Then $z \in E \cap (B(x,r) \setminus \{x\})$ which completes the proof.