I am trying to prove that $$V=\{(x,y) \in \mathbb{C}^2| \ x^2+y^2 = 1\} \simeq \mathbb{C}^* = \mathbb{C}\setminus \{0\}$$ where $\simeq$ is to be intended as homeomorphic.
Fix $y\neq \pm 1$ then $x^2 = 1-y^2$ gives me two distinct solutions for $x$. It follows that $V$ will contain two disjoint copies of $\mathbb{C}\setminus \{\pm 1\}$.
Now it remains to see if there is any solution of the form $(x,\pm 1)$. There are exactly two solutions $\{(0, \pm 1)\}$. Then it seems reasonable to me that $V$ is obtained filling the two points in the two copies of $\mathbb{C}\setminus \{\pm 1\}$ and identifying the points $\pm 1$, more explicitely $$V = \frac{(\mathbb{C}\setminus \{\pm 1\} )\times \{a,b\}}{\sim}$$ where $(+1,a)\sim (+1,b)$ and $(-1,a) \sim (-1,b).$
The problems is that this is not the cylinder $\mathbb{C}^*$, indeed if we remove the two points $[(\pm 1,a)]$ we disconnect $V$ while $\mathbb{C}^*$ remains connected.
what is wrong?
Your approach shows the following:
We have a map $V \to \mathbb{C}$ given by $(x, y) \mapsto y$. This map is surjective. The fiber over every point other than $y= \pm 1$ has size two, and the fiber over the two points $y = \pm1$ has size one.
You have concluded that away from $\pm 1$ this is the trivial (disconnected) double cover, and the rest is obtained by a gluing process. But even before we get to the gluing part, the part about the trivial double cover is already not true; for example, take $y = 0$. Then we can connect the two pre-images of $y$ in $V$ by an arc in $V$ (just take a path from $x=1$ to $x=-1$ in $\mathbb{C}$ which avoids $x=0$).
For a different approach, see the comment I left on the question.