Fundamental group of a sphere with straight line joining the north and south poles This problem comes from the book Topology and Geometry of Bredon :

Let $X$ be the union of the unit sphere in 3-space with the straight line segment from the north pole to the south pole. Find $\pi_1(X)$.

I am reading Atcher's book at the same time. He gives a CW-complex structure to this space, and since the line is contractible, he identifies this space to $S^2 \vee S^1$ from which we can compute the fundamental group using Seifert-van Kampen's theorem and we get
\begin{equation*}
\pi_1(X) \approx \mathbb{Z}.
\end{equation*}
Nevertheless, one should be able to use Seifert-van Kampen's theorem directly to this space without relying on some CW-complex structure as it is not yet introduced in the book of Bredon at this stage.
I have tried the following :
Consider the open cover $\{U, V\}$ where $U$ is a thickened neighborhood of the union of the north hemisphere and the middle-top part of the straight line, and where $V$ is a thickened neighborhood of the union of the south hemisphere and the middle-bottom part of the straight line. But, the way I understand it, they are both contractible (they contract to the north and south pole respectively), and thus the fundamental group vanishes by Seifert-van Kampen's theorem.
Where is the error in this argument?
 A: Consider $U$ and $V$ to be semispheres, i.e. $U$ goes from the north pole to a bit past the equator, and $V$ from a bit above the equator to the south pole. But don't imagine the line to be perpendicular to the equator plane but rather contained in it. Or if you will, construct $U$ and $V$ relative to Greenwich's meridian :)
If you stare at $U$ and $V$ for a while, they are (homotopic to) disks with a cord attached to two points of the boundary. Collapsing the disks, one gets moreover a homotopy equivalence with $S^1$ in both cases.
The intersection is now a band along the equation together with the line. Via a deformation retraction, this is homotopy equivalent to just the equator - i.e. $S^1$ - and the line, which is like a figure $8$ (collapse the line).
Hence by van Kampen we have $\pi_1(X) = \pi_1(U) \ast \pi_1(V) / \langle \langle  \iota_1(\omega)\iota_2(\omega)^{-1} : \omega \in \pi_1(U \cap V)\rangle\rangle$, and $\pi_1(U) = \pi_1(V) = \Bbb Z$.
The intersection $U \cap V$, as we have said, is homotopy equivalent to a figure $8$, whose fundamental group is the free group on two generators, one for each loop. Chasing the homotopies, we obtain that the generators for $\pi_1(U \cap V)$ correspond to going through the line and then to one half of each semicircle of the equator. If $a$ and $b$ are the generators of $\pi_1(U)$ and $\pi_1(V)$, then these loops correspond to $a$ and $b$ so the relation introduced is $a = b$.
Finally, this gives $\pi_1(X) = \langle a,b : a = b\rangle \simeq \Bbb Z$, as desired. Maybe a different choice of orientations changes the presentation a little bit (e.g. the relation could be $ab = 1$, which still yields the correct group).
