This question may be trivial, I did not think hard about it. 


A friend of mine is looking for an irreducible (reduced) analytic subspace $C \subset \mathbb{C}^2$ with the following property. Let $f \colon C \rightarrow \mathbb{C}$ be the projection on the first factor. He wants that


1) All singular points of $C$ and all ramification points for $f$ lie in a limited set, so removing that set we obtain a topological covering from some open set of $C$ to $\mathbb{C}$ with a ball removed.


2) That covering should be trivial (even better if it is finitely-sheeted).


So the curve $C$ is connected, but only if one passes near the origin. Sufficiently far from that ther should be no way to jump between sheets. Is it possible to find such a $C$?



Riemann surface disconnected at infinity
This is an extended version of my comment. Suppose we stay on the surface $z^2+w^2=1$ but away from the origin. The identity
$|z^2+w^2|^2=|z|^4+|w|^4+2Re((z \bar w)^2)$ 
tells us that the square of $z \bar w$ has negative real part. The set of complex numbers $\zeta$ such that $Re(\zeta^2)<0$ has two connected components: it's the disjoint union of two open sectors. Finally, note that switch from (z,w) to (-z, w) involves going from one component to the other. 


