Help finding the fundamental group of $S^2 \cup \{xyz=0\}$ let $X=S^2 \cup \{xyz=0\}\subset\mathbb{R}^3$ be the union of the unit sphere with the 3 coordinate planes. I'd like to find the fundamental group of $X$.
These are my ideas:
I think the first thing I should do is to retract all the points outside the sphere to the sphere (is that possible? how?)
then using spherical coordinates I could make the following deformation: 
$(1,\phi,\theta)\to ((\sin\phi \sin\theta \cos \phi\cos \theta)^t,\phi,\theta)$. This collapses all the point in $S^2 \cap \{xyz=0\}$ to $0$ obtaining $8$ deformed spheres touching each other in $0$ (how can I prove rigorously that they are simply connected), using Van Kampen theorem we can say that $X$ is simply connected.

If you want to pursue the argument you started (which seems fine to me), I suggest that you consider the map 
$$
F: R^3 \times I  \to R : (x, s) \mapsto \begin{cases}
x & |x| \le 1 \\
(1-s) \frac{x}{\|x\|} + s (x - \frac{x}{\|x\|}) & \text{otherwise} 
\end{cases}
$$
That retracts things onto the sphere plus the three coordinate disks within it. 
If you now let $U'$ be the exterior of a ball of radius $1/2$, and $V'$ be the interior of a ball of radius $3/4$ in 3-space, and $U$ and $V$ be the intersections of these with your space, you can apply van Kampen and find that $\pi_1$ is trivial. 
