Understanding the theorem:
Let's consider the metric space $\Bbb{Q}$ with the euclidian metric from $\Bbb{R}$.
Let $a,b\in$ $\Bbb{R}\setminus{\Bbb{Q}}$ with $a<b$ and $S=(a,b)\cap\Bbb{Q}$ . Prove that $S$ is closed and bounded in $\Bbb{Q}$, but not compact.
My question is why $S$ is not compact if Heine-Borel's theorem tells me that if a set is closed and bounded then the set is compact.
Heine-Borel's theorem talks about subsets of $\mathbb{R}^n$ but you are dealing with a subset of $\mathbb{Q}$. While being bounded doesn't depend on the choice of the "big" space being closed does. For example $(-\sqrt{2}, \sqrt{2})\cap\mathbb{Q}$ is closed in $\mathbb{Q}$ but not in $\mathbb{R}$.
So back to your question. For any real number $r\in\mathbb{R}$ there exists a sequence $(q_n)\subseteq\mathbb{Q}$ convergent to $r$. Also if $a,b\in\mathbb{R}$, $a<b$ then there exists a irrational $z\in\mathbb{R}\backslash\mathbb{Q}$ such that $a<z<b$. These two facts together imply $(a,b)\cap\mathbb{Q}$ cannot be compact for any $a<b$.
Now your $S$ is obviously bounded. It is also closed because $(a,b)\cap\mathbb{Q}=[a,b]\cap\mathbb{Q}$ if $a,b$ are irrational.