TITLE: Closed and bounded does not mean compact in general
QUESTION [5 upvotes]: Suppose in metric space $\Bbb Q$ I consider the subset $\{x\in\Bbb Q | 2<x^2<3\}$. I want to find a open cover for this set which does not have any finite subcover. How to do it ? Hint is enough.

REPLY [2 votes]: Hint: Consider a cover consisting of $\Bbb{Q} \cap (-r, r)$ for rationals $r$ with $r < \sqrt{3}$.