Show that set $A=${$x \in \mathbb{Q} :-1<x<1$} is open in $\mathbb{Q}$ but not closed in $\mathbb{Q}$
The set $A$ is open in $\mathbb{Q}$ as we take any point $p \in \mathbb{Q}$ and $p \in A$ then we can choose an $ \epsilon >0$ then $p - \epsilon >-1$ and $p + \epsilon <1$,then we see that $ \epsilon < 1$, we can actually get a neighborhood such that set is open.
We see that $1$ is also a limit point of $A$ but $1$ doesn't belong to the set so $(A)^{'} \ne A$ hence these is not closed . Is this attempt correct
Yes, that's correct. Another way to look at it is: $A$ is closed in $\mathbb{Q}$ if and only if there is a closed set $F$ in $\mathbb{R}$ such that $F \cap \mathbb{Q} = A$. So in particular, $A \subseteq F$ and hence the closure of $A$ is a subset of $F$. But what is the closure of $A$ in $\mathbb{R}$? And then what is $\bar{A} \cap \mathbb{Q}$?