Set of points of continuity are $G_{\delta}$
The definition of $A_n$ is a bit confusing. I think that what you want here is to let
$$A_n=\left\{x\in\Bbb R:\exists r_n(x)>0\,\forall x',x''\in B\big(x,r_n(x)\big)\left(\left|f(x'')-f(x')\right|<\frac1n\right)\right\}\;.$$
You definitely want the absolute values, and you need to say that it’s the points $x$ for which such a neighborhood $B\big(x,r_n(x)\big)$ exists. You don’t have to indicate explicitly the dependence of $r$ on $n$ and $x$ as I did here, but it doesn’t hurt, especially when you’re learning.
Now let $G=\bigcap_{n\in\Bbb Z^+}A_n$, and let $C=\{x\in\Bbb R:f\text{ is continuous at }x\}$. You need to show three things:
- Each $A_n$ is open.
- $C\subseteq G$. This is your ‘$\Rightarrow$’.
- $G\subseteq C$. This is your ‘$\Leftarrow$’.
You omitted (1) altogether, but it’s not hard: just show that if $x\in A_n$, then $B\big(x,r_n(x)\big)\subseteq A_n$, and conclude that $A_n=\bigcup_{x\in A_n}B\big(x,r_n(x)\big)$ and hence is open.
You’ve essentially got (2), but it could be stated much more clearly. Suppose that $x\in C$ and $n\in\Bbb Z^+$. Then there is an $r_n(x)>0$ such that $|f(x')-f(x)|<\frac1{2n}$ for all $x'\in B\big(x,r_n(x)\big)$. But then by the triangle inequality $$|f(x'')-f(x')|\le|f(x'')-f(x)|+|f(x)-f(x')|<\frac1n$$ for all $x',x''\in B\big(x,r_n(x)\big)$, so $x\in A_n$. And since $n\in\Bbb Z^+$ was arbitrary, $x\in G$.
Much the same applies to (3). Suppose that $x\in G$, and let $\epsilon>0$ be arbitrary. There is an $n\in\Bbb Z^+$ such that $\frac1n\le\epsilon$, and $x\in A_n$, so $|f(x')-f(x)|<\frac1n\le\epsilon$ for all $x'\in B\big(x,r_n(x)\big)$, i.e., for all $x'$ such that $|x'-x|<r_n(x)$, and it follows immediately that $x\in C$.