Is the set $\left\{x \in Q\colon x^2 \le 2\right\}$ open or closed? Is the following set open or closed? I am almost certain it is open as the limits are not included in the rational set.
$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\}$$
What I really don’t understand is the proper closure of the set. I think it would be:
$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\} \cup \left\{\pm\sqrt{2}\right\}$$
But then again, the following seems reasonable (although not an “efficient” closure):
$$\left\{x \in \mathbb{R}\colon x^2 \le 2\right\} = \left[-\sqrt{2},\sqrt{2}\right]$$

Let $A$ be your set of rationals.  You ask if $A$ is open or closed.  
Open or closed within what space? The reals $\mathbb{R}$ or the rationals $\mathbb{Q}$?
Notice that $$A = [-r,r] \cap \mathbb{Q} = (-r,r) \cap \mathbb{Q}$$
where r = sqrt 2.  Thus within $\mathbb{Q}$, $A$ is clopen.
However,  within $\mathbb{R}$ it is neither.  
The $\mathbb{R}$-closure of $A$ is $[-r,r]$.  The $\mathbb{R}$-interior of $A$ is empty.  
