Does a bounded helix; for instance $\{(\cos 2\pi t, \sin 2\pi t, t); -5\leq t\leq5\}$ in $\mathbb R^3$ with the projection map $(x,y,z)\mapsto (x,y)$ form a covering space for the unit circle $\mathbb S^1$?
If yes, does it disprove: There is a bijection between the fundamental group of a space and the fibers of the base point into a simply connected covering space?
Yes it's a covering map. Since the fundamental group of $S^1$ is $\mathbb{Z}$, and since each fiber is the set of $(\cos(t_0),\sin(t_0),t_0+2\pi k)$, $k \in \mathbb{Z}$, there is no conflict to the bijection you mention.
Note the question has been reformulated and this answer is no longer relevant; I'm awaiting further clarification from OP as to what the rephrased bijection assertion means.
The re-edit now has the parameter $t$ bounded, i.e. $t \in [-5,5]$, in the helix parametrization $(\cos 2\pi t,\sin 2\pi t, 2\pi t).$ This is not a covering space, since for $t$ near an end, say near 5, the neighborhood of $(0,0)$ in the base is not covered by an open interval along the helix. In a simple covering space over the circle, if you are on a sheet of the cover over any point in the base, you should be able to move back or forth a bit and remain on that sheet, locally.