To prove an if and only if, we need to prove the $\implies$ direction and the $\impliedby$ direction.
For the $\implies$ direction, suppose $A \subseteq \mathbb{R}$ is closed.
We want to prove then that if $\{ x_{n} \}_{n = 1}^{\infty}$ is a sequence of real numbers in $A$ that converges, then its limit is in $A$. If we call $x$ the limit of this sequence (i.e., $\lim x_n = x$), then by definition of convergence, given $\epsilon > 0$, there exists an $N$ such that if $n \geq N$ then $d(x_{n}, x) < \epsilon$ (alternatively, for all $n \geq N$, $x_{n} \in B(x, \epsilon)$).
To show that every convergent sequence contained in $A$ has its limit in $A$, suppose by contradiction that there is a convergent sequence $\{ x_{n} \}_{n = 1}^{\infty}$ in $A$, but its limit, $x$, is in $\mathbb{R} \setminus A$. Since $A$ is closed, $\mathbb{R} \setminus A$ is open. Since $x \in \mathbb{R} \setminus A$, and the set is open, we know by definition of open that $\exists \epsilon > 0$ such that $B(x, \epsilon) \subseteq \mathbb{R} \setminus A$. So we found a ball around $x$ entirely contained in $\mathbb{R} \setminus A$. But by definition of a convergent sequence, $\exists N$ such that for all $n \geq N$, $x_{n} \in B(x, \epsilon)$. So there is a point in the sequence after which all terms are in $B(x, \epsilon)$. But $B(x, \epsilon) \subseteq \mathbb{R} \setminus A$. Which means there are points of the sequence in $\mathbb{R} \setminus A$. This contradicts the assumption that the sequence was entirely contained in $A$. Thus, every convergent real sequence contained in $A$ has its limit in $A$, as desired.
Hopefully you should now have some idea on how to prove the other direction. Just remember your definitions.
Best Answer
OK, you are immediately doing something wrong here: You need to show that A∪B = S, so you should not assume it.
The basic set-up you should use is to show two things:
First assume that $X\in A \cup B$. Now show that $X\in S$. This shows $A\cup B \subseteq S$
Now assume that $X\in S$, and show that $X\in A \cup B$. This shows $S \subseteq A\cup B$
Of course, in both cases, use the fact that $A^C \subseteq B$
Once you have shown these two things, you have shown that $A\cup B = S$