Suppose $f:R\to\,R$ is a function. For $k\in\mathbb{Z}^{+}$, let $G_k=\{a\in\,R\,:\,\text{there exists a $\delta$>0 such that $|f(b)-f(a)|<1/k$ for all $b,c\in(a-\delta,a+\delta)$}\}$ Prove that the set of points at which $f$ is continuous equals $\cap_{k=1}^{\infty}G_k$

should be $|f(b)-f(c)|<1/k$

is $G_k$ a open/close set?

Let $I_k=(a-\epsilon_k/2,a+\epsilon_k/2)$, we then have $G_k\subset\cup_{k\in\,K}I_k$

so $G_k$ is open, not sure how to show the required statement

This is often used in the proof that the set of continuity points is $G_\delta$, so checking some sources about that might be useful

On the main site you can find How to show that the set of points of continuity is a $G_{\delta}$ or Set of points of continuity are $G_{\delta}$.

@Simple There is something wrong with this. (Or at least the way it is written.) It is not clear to me what you mean by $\epsilon_k$. But regardless of that $G_k$ does not depend on $a$, but the $\bigcup_{k\in K} I_k$ depends on $a$. (Since your definition of $I_k$ depends on $a$.)

Claim. $f$ is continuous at $a$ $\Leftrightarrow$ $a\in\bigcap G_k$

$\boxed\Rightarrow$ Let us fix some $k$ and choose $\delta>0$ such that $$|f(x)-f(a)|<\frac1{2k}$$ for $|x-a|<\delta$.

Then for $b,c\in(a-\delta,a+\delta)$ we have $$|f(b)-f(c)|\le |f(b)-f(a)|+|f(a)-f(c)| < \frac1{2k} + \frac1{2k} = \frac1k.$$

This shows that $a$ belongs to $G_k$.

Since this is true for every $k$, we have shown that $a\in\bigcap G_k$.

$\boxed\Leftarrow$ Let us assume that $a\in\bigcap G_k$.

Suppose that $\varepsilon>0$ and choose $k$ such that $\frac1k<\varepsilon$.

In particular, if $|a-x|<\delta$, then $a,x\in(a-\delta,a+\delta)$ and we get $$|f(x)-f(a)|<\frac1k<\varepsilon.$$

This proves that $f$ is continuous at $a$.