An open covering of $\mathbb{Q} \cap [0,1]$ that does not contain any finite subcovering

Consider the topological subspace $\mathbb{Q} \cap [0,1]$ endowed with the usual topology of $[0,1]$, since $[0,1]$ is Hausdorff and that $\mathbb{Q} \cap [0,1]$ is not closed, we conclude that $\mathbb{Q} \cap [0,1]$ is not compact, i.e., there exist an open covering $\{O_i\}_{i \in I}$ of $\mathbb{Q} \cap [0,1]$ such that for any finite index set $J \subseteq I$, $\displaystyle \bigcup_{j \in J} O_j \subset \mathbb{Q} \cap [0,1]$, where the inclusion is strict.

My question is if there is an explicite example of such open covering?

Pick your favorite irrational number $\xi\in (0,1)$, and consider the open cover $$\Big\{\Big[0,\xi-\frac{1}{n}\Big)\cup\Big(\xi+\frac{1}{n},1\Big]\Big\}_{n=n_0}^{\infty}$$ where $n_0$ is chosen large enough that $\xi-\frac{1}{n_0}>0$ and $\xi+\frac{1}{n_0}<1$.