Show that geodesic distance on the $n$-sphere respects the triangle inequality

You are projecting $x, z$ onto $y$ respectively: Consider the plane spanned by $x, y$ (let's say $x\neq \pm y$). Then one can find a unit vector in this plane which is orthogonal to $y$. Call this $u$. Then one has

$$ x = \langle x, y\rangle y + \langle x, u\rangle u.$$

By definition, $\arccos \langle x, y\rangle = \theta$, so $\cos\theta = \langle x, y\rangle$. Since$x, y, u$ are orthonormal to each other, this force $\langle x, u\rangle = \sin \theta$. Thus

$$ x = y\cos \theta + u\sin\theta .$$

(Similar for $z,v$).

(2) is really easy calculations, starting at

$$ \langle x,z\rangle = \langle y \cos\theta + u\sin\theta , y\cos\varphi + v\sin\varphi \rangle$$

and then expanding the right hand side.

Given three points on the $n$-sphere, there is a $3$-plane containing them and the origin; the intersection of this with the sphere is totally geodesic.

It suffices to prove things for $S^2,$ by whatever means you like.