I want to show that $(U, \phi=(\lambda,\theta))$ is a smooth coordinate chart for the standard smooth structure of $\mathbb{S}^2$, where $\lambda \in (-\pi/2, \pi/2)$ is the latitude, $\theta \in (0, 2\pi)$ is the longitude and $U = \mathbb{S}^2 \setminus\{0-meridian\}$.
Proof. Consider $F \colon (-\frac{\pi}{2},\frac{\pi}{2})\times (0,2\pi) \to \mathbb{R}^3$ defined as $$F(\lambda,\theta)=(\cos\theta\cdot\cos\lambda, \sin\theta\cdot\cos\lambda, \sin\lambda).$$ It is easy to show that $Im(F) = U$ and that $F$ is injective so $F|^U \colon (-\frac{\pi}{2},\frac{\pi}{2})\times (0,2\pi) \to U$ is a bijective map: this map is also smooth because $U$ is an embedded submanifold of $\mathbb{R}^3$ so we can restrict the image and retain smoothness.
We have that $F|^U$ is a full rank map because $F$ is and $F=\iota \, \circ F|^U$ ($\iota$ is smooth embedding) so $F|^U=\phi^{-1}$ is a diffeomorphism from the rectangle $(-\frac{\pi}{2},\frac{\pi}{2})\times (0,2\pi)$ to $U$. In fact bijective + smooth + full rank implies diffeomorphism. $\square$
Is this proof ok? Another way to proceed could have been: show $(U, \phi)$ is a coordinate chart and then prove that is smoothly compatible with a smooth atlas for $\mathbb{S}^2$, maybe with the atlas given by the stereographic projections. But I find this more long and tedious.