I have proved the following statement(s) and I would like to know if my proof is correct and/or/how it could be improved, thank you.
Suppose $f:\mathbb{R}\to\mathbb{R}$ is a function.
(a) For $k\in\mathbb{Z^+}$, let $G_k=\{a\in\mathbb{R}:\text{there exists }\delta>0\text{ such that }|f(b)-f(c)|<\frac{1}{k}\text{ for all }b,c\in (a-\delta,a+\delta)\}$.
Prove that $G_k$ is an open set for each $k\in\mathbb{Z^+}$;
(b) Prove that the set of points at which $f$ is continuous equals $\bigcap_{k=1}^{\infty}G_k$
(c) Conclude that the set of points at which $f$ is continuous is a Borel set.
(a) Let $k\in\mathbb{Z^+}$ and $\alpha\in G_k$: then there exists $\delta>0$ such that $|f(b)-f(c)|<\frac{1}{k}$ for all $b,c\in (\alpha-\delta,\alpha+\delta)$.
Now, let $\delta_{\alpha}:=\frac{\delta}{4}$ and take $\beta\in B_{\delta_{\alpha}}(\alpha):=(\alpha-\delta_{\alpha},\alpha+\delta_{\alpha})\subset (\alpha-\delta,\alpha+\delta)$: if we let $\delta_{\beta}:=\frac{\delta_{\alpha}}{4}$ and take $b,c\in B_{\delta_{\beta}}(\beta):=(\beta-\delta_{\beta},\beta+\delta_\beta)\subset B_{\delta_{\alpha}}(\alpha)$ we have that $|f(b)-f(c)|<\frac{1}{k}$ thus $B_{\delta_\alpha}(\alpha)\subset G_k$ ie we have found an open ball centered in $\alpha$ contained in $G_k$ which is thus open, as desired.
(b) Let $x_0\in\mathbb{R}$ be a point where $f$ is continuous and $k\in\mathbb{Z^+}$: then by definition of continuous function for every $\varepsilon>0$ there exists $\delta_{\varepsilon}>0$ such that $|f(x)-f(x_0)|<\varepsilon$ for all $x\in (x_0-\delta_{\varepsilon},x_0+\delta_{\varepsilon})$ so in particular (since $\frac{1}{k}>0\ \forall k\in\mathbb{Z^+}$) for every $k\in\mathbb{Z^+}$ there exists $\delta_{k}>0$ such that $|f(x)-f(x_0)|<\frac{1}{2k}$ for all $x\in (x_0-\delta_{k},x_0+\delta_{k})$ and if we take $b,c\in(x_0-\delta_{k},x_0+\delta_k)$ we have that $|f(b)-f(a)|=|f(b)-f(x_0)+f(x_0)-f(a)|\leq |f(b)-f(x_0)|+|f(x_0)-f(a)|<\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}$ so $x_0\in G_k$ for every $k\in\mathbb{Z^+}$ thus $x\in\bigcap_{k=1}^{\infty}G_k$.
Now let $x_0\in\bigcap_{k=1}^{\infty} G_k$ and $\varepsilon>0$: then $x\in G_k$ with $k>\frac{1}{2\varepsilon}$ so there exists $\delta_k$ such that $|f(b)-f(c)|<\frac{1}{2k}<\varepsilon$ for all $b,c\in (x_0-\delta_k,x_0+\delta_k)$ so it is also $|f(b)-f(x_0)|=|f(b)-f(c)+f(c)-f(x_0)|\leq |f(b)-f(c)|+|f(c)-f(x_0)|<\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}<\varepsilon$ for all $b\in (x_0-\delta_k,x_0+\delta_k)$ so $f$ is continuous at $x_0$.
Thus $\{x\in\mathbb{R}:f\text{ is continuous at }x\}=\bigcap_{k=1}^{\infty}G_k$.
(c) Since from point (b) we know that $\{x\in\mathbb{R}:f\text{ is continuous at }x\}=\bigcap_{k=1}^{\infty}G_k$ and from point (a) that the $G_k$s are open sets in $\mathbb{R}$ hence Borel sets, since $\sigma$-algebras like the set $\mathcal{B}$ of Borel sets are closed under intersections, we have that $\bigcap_{k=1}^{\infty} G_k$ is a Borel set hence $\{x\in\mathbb{R}:f\text{ is continuous at }x\}$ is a Borel set too.