Prove that the helicoid is homeomorphic to the plane.
Good!
Here are a few abstract thoughts.
We know the universal covering space of the punctured plane $\displaystyle \mathbb{C}-\{0\}$ is the plane $\displaystyle \mathbb{C}$, with the projection $\displaystyle \mbox{exp }: \mathbb{C} \to \mathbb{C}-\{0\}$.
On the other hand, the universal cover of $\displaystyle \mathbb{C}-\{0\}$ is also the Riemann surface of the (multi-valued) inverse function $\displaystyle \mbox{Log}$, of which the helicoid is the most obvious model. By the universal property of the universal cover, there exists a homeomorphism between the helicoid and the plane.
(Informally : the slit plane $\displaystyle \mathbb{C}-\mathbb{R}_{\geq 0}$ is easily seen to be homeomorphic to a strip without its boundary, (any branch of the logarithm doing the job). Now lift the cut to the helicoid, and map each slice of the helicoid to a strip, identifying the appropriate boundaries of two strips if the corresponding boundaries of the slices are identified on the helicoid. In this way, the helicoid is mapped homeomorphically on the plane.)