$S^2$ is not a countable union of embedded circles $S^1$

Topological dimension is a nice invariant, especially in separable metric spaces:

Embedded copies of $S^1$ will be nowhere dense in $S^2$ (it cannot have interior in $S^2$ as $S^1$ has dimension $1$), so Baire's theorem shows 1.

The countable sum theorem in dimension theory implies 2. easily.

To add to Henno's nice answer let me point out that you may use homology (the invariance of domain theorem) if for some reason you prefer that to some elementary dimension theory to see that embedded circles are nowhere dense. You can also use measure theory for problem #2 but not for problem #1, as there are Jordan curves with nonzero measure.