Given the advertised starting decomposition of the lower closed unit disk, each 0-cell is a subset of [-1,1]. The union of the 0-cells is typically dense G-delta, and has measure 2, if we use Lebesque measure on the Euclidean interval [-1,1].

I think your question is a good one, since in general the closure of a compact subspace can fail to be compact, if the spaces in question fail to be Hausdorff. But in this case the complement of X is finite, so all is well, the closure of X is compact.

Yes, given arbitrary open cover, one of the open sets U contains z. The complement of U is closed in the compact space [0,1], and hence can be covered by finitely many of the surviving open sets in the original cover.

Let U be an arbitrary open dense subspace of [0,1], with the relative topology. For example U=[0,1/2) union (1/2,1). The set z union U is a typical open set in X which contains z.

Isn;t F torsion free? Thus g in F has infinite order if g is not id.