Let $\displaystyle A = \mathbb{Q} \cap [0,1] \subset \mathbb{R} $. Give an example of an open cover of $\displaystyle A $ which has no finite subcover.

I was thinking, since it specifies exactly that $\displaystyle A \subset \mathbb{R} $, then do our open sets (that are in the open covers) have to be open in $\displaystyle \mathbb{R} $, or can they be open in $\displaystyle \mathbb{Q} $?

For example, if $\displaystyle x $ is any rational number in $\displaystyle [0,1] $ then the set $\displaystyle \{ x \} $ is open in $\displaystyle \mathbb{Q} $, and the collection of all sets {x} such that x is a rational number in [0,1] is an open cover which has no finite subcover. But is it permissible for my open sets to be open in $\displaystyle \mathbb{Q} $? Or do they have to be open in $\displaystyle \mathbb{R} $?