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By replacing each set with its complement, this problem is equivalent to this one: en.wikipedia.org/wiki/Set_cover_problem. So it's NP-complete. You can do a lot better though if you're happy with an approximate solution.
... and every finitely generated abelian group is the additive group of some associative unitary commutative ring, by looking at the classification: en.wikipedia.org/wiki/…. Is there any (necessarily not finitely generated) abelian group which is not the underlying additive group of some commutative ring?
Unfortunately, $[0,1]\backslash(\mathbf{Q}\cap[0,1])$ is also completely metrisable: take the distance to be one over the first index at which the continued fraction expressions differ.
Intuitively, we may deform $f$ so that on its initial anticlockwise run it is very slightly nearer the origin than described above, then it smoothly but sharply turns right to do its clockwise run, then it smoothly but sharply turns left to do its final anticlockwise run. All this is done so that $f$ stays in a very small neighbourhood of $S^1$. Finally, just before $f$ returns to its starting point, it has a bit of freedom and so may behave however it must so that $a_1=0$ in the end: this will involve either cutting the corner slightly or ballooning out slightly.
