The line x=1 in the (x,y) plane. The line (1,y,0) in (x,y,z) space.

No, the spin I describe yields a compactification of the open unit disk by attaching a cylinder.

No, the 0th homotopy group of a non path connected space is nontrivial. If this doesn't count, spin the original graph about the line x=1 in (x,y,z) space. Then X will have nontrivial fundamental group/

David's counterexample is correct. Every open subset of the plane has free fundamental group with at most countably many generators.

To fill in a few details in the above, suppose K and K' are nonempty compact totally disconnected subsets of the 2-sphere, with respective complements U and U'. Then any homeomorphism between U and U' extends to a homeomorphism of the 2-sphere, and any homeomorphism between K and K' extends to a homeomorphism of the 2-sphere. For the latter direction, see the pf. in Van Mill's book on infinitely dimensional topology that the Cantor set cannot be wildly embedded in the plane. For the former direction apply the Schoenflies theorem repeatedly.

Thanks Sergei! My obtuse (and likely incorrect) definition should be ignored. The reader should use any standard definition of length space. For example, for all x and y in X and all e>0, there exists a rectifiable path from x to y whose length is less then d(x,y)+e.