The following two are from Introductory Functional Analysis by E Kreyszig:
2.5-2 Lemma . A [sequentially] compact subset M of a metric space is closed and bounded.
The converse of this lemma is in general false.
However, for a finite dimensional normed space we have :
2.5-3 Theorem. In a finite dimensional normed space X, any subset $M \subset X$ is [sequentially] compact if and only if M is closed and bounded.
On the other hand, here it shows that $\mathbb{Q}\cap[0,2]$ is not sequentially compact thus must be infinite dimensional (over $\Bbb R$ or $\Bbb C$) but any element of $\mathbb{Q}\cap[0,2]$ can be written as a multiple of some element in $\Bbb R$ (or $\Bbb C$) times 1 so dimension of $\mathbb{Q}\cap[0,2]$ over $\Bbb R$ (or $\Bbb C$) is 1 not infinite! Contradiction?
Since $\mathbb Q\cap[0,2]$ is not a normed space, there is no contradiction.
On the other hand, $\mathbb R$ is a $1$-dimensional normed space and, indeed, $\mathbb Q\cap[0,2]$ is a subset of $\mathbb R$ whichnot closed and not compact. So, again, there is no contradiction.