Cantor set--nowhere dense, complete I can't figure out this out.

Cantor set is closed in $\mathbb{R}$.

$\mathbb{R}$ is a complete metric space.

Every closed subset of a complete space is also complete; thus, so is the Cantor set.

Complete space can't be written as a countable union of nowhere dense sets; thus, the Cantor set has no such representation. (1)

Cantor set is nowhere dense (its interior is empty, and no intervals are contained in it)

Thus, the Cantor set is a finite (then countable) intersection of nowhere dense sets. 

Contradiction with statement (1).

Where is a mistake? 

Does representation in (1) refer only to infinite countable intersections?
the assumption that since its finite thus countable is false.  Countable means a bijection with N.  It can be infinite (countably infinite, or simply just stated countable)
the assumption that since its finite thus countable is false.  Countable means a bijection with N.  It can be infinite (countably infinite, or simply just stated countable)