$A =\{x\in\Bbb Q\mid 0≤x^2

$A =\{x\in\Bbb Q\mid 0≤x^2<2\}$. Is the closure of $A˚$ compact?

I think that it is, because if I take the closure of the interior of A, I'm adding every limit point on A, for example sqrt(2), so the sub sequences will all converge within the set. If my argument is correct, its not valid if $0<x$ strictly right? because one would never achieve the value zero? Or having the closure of the set automatically adds it?