How can Clusters in Cantor dust absorb enough irrational points? [on hold]

Question

Let all rational numbers $q_n$ of the real axis be covered by intervals $I_n$ of measure $|I_n| = 2^{−n}$, such that $q_n$ is the centre of $I_n$. Then the endpoints are rational numbers. The irrational numbers $x$ of the complement of infinite measure, not covered by the intervals $I_n$, form particles of a totally disconnected space, so-called "Cantor dust" $\Xi$.

Every particle $x\in\Xi$ must be separated from every particle $y\in\Xi$ by at least one rational number $q_n$ and hence by at least one interval $I_n$ covering $q_n$. Since the end points of the $I_n$ are rational numbers too, also being covered by their own intervals, the particles of Cantor dust can only be limits of infinite sequences of overlapping intervals $I_n$.

If intervals don't overlap, then they cannot form a common limit; their limits must lie between them. But in any case infinitely many finite intervals are required to establish one limit. And in any case two limits must be separated by infinitely many overlapping intervals

Such an infinite set of overlapping intervals is called a cluster. In principle, given a fixed and complete enumeration of the rationals, we can calculate every cluster and the limits of its union. Therefore, every irrational $x\in\Xi$ can be put in bijection with the pair of clusters, i.e., the infinite set of intervals converging to it. There are countably many sets $I_n$ and therefore not more than countably many disjoint clusters with limits and therefore not more limits.

Where are the other irrational numbers of the complement?

EDIT: Maybe that the closers thought to have the Cantor-sets as a counterexample. While Cantor-sets are based on the same fallacy of set theory, namely the unjustified extrapolation from finite to infinite bijections, they are not as easily to refute because there is not the crucial condition that irrational points must be separated by clusters of finite measure.