One could easily show that all $x\in \mathbb{R}$ is a Borel set, by simply noticing that $$x=\bigcap_{j\in\mathbb{N}} (x-\frac 1 j,x+\frac 1j)$$
is the countable intersection of Borel sets, and since the Borel $\sigma$-algebra is $\cap$-stable the statement follows.
And since we know that the Borel $\sigma$-algebra is stable under countable unions (intersections), I don't seem to understand how the Cantor set can have subsets that are not Borel sets.
I've seen numerous answers in the site proving that statement by using the fact that the cardinality of the Cantor set is $2^{\aleph_0}$ (a…