Why can't Cantor sets cover $\mathbb{ R}$?

You can't cover a complete metric space with a countable union of closed nowhere dense subsets, by Baire's theorem. Cardinality tells us very little. Even infinite dimensional but separable Banach spaces have the same cardinality as lines, but perhaps here it is more intuitive that you will never cover such a space with countably many lines. Similarly, you can't cover the plane with countably many lines, polygons, conic sections, etc.

To see how Baire's theorem is more fundamental than measure here, note that the same is true for "fat" Cantor sets. That is, you take out smaller intervals to obtain Cantor-like sets with positive measure. These sets are still closed and nowhere dense, so by Baire's theorem the line is not a countable union of such sets, even though comparing measures wouldn't tell you this.

It's not just that the Cantor set has measure 0, it's also that the reals are a Baire space and the Cantor set is nowhere dense. E.g. take a fat Cantor set, $C$, and look at $C + \mathbb{Z}$ which has infinite measure but is still be nowhere dense.

UPDATE: There is a related discussion on MO that could be of interest.