How to find the Riemannian Distance on the Sphere $S^n\subset \Bbb{R}^{n+1}$

Let $p=(1,0,\dots,0)$ and $q=(-1,0,\dots,0)$. Then any path from $p$ to $q$ is of the form $\gamma(t)=(t,h(t))$, where $h(t)\in \Bbb{R}^n$. Then the length of the path is $$ L=\int_{-1}^{1}|\gamma'(t)|dt. $$ But $\gamma'(t)=(1,h'(t))$, and so $|\gamma'(t)|=\sqrt{1+|h'(t)|^2}$.

Now $t^2+\langle h(t),h(t)\rangle=\langle \gamma(t),\gamma(t)\rangle=1$, so $$ 2t+2\langle h(t),h'(t)\rangle=0. $$ By Cauchy Schwartz we have $|h'(t)|^2|h(t)|^2\ge |\langle h(t),h'(t)\rangle|^2=|t|^2$, and so $$ 1+|h'(t)|^2\ge \frac{t^2}{|h(t)|^2}+1, $$ but $|h(t)|^2=1-t^2$, hence $$ \sqrt{1+|h'(t)|^2}\ge \sqrt{1+\frac{t^2}{1-t^2}}=\sqrt{\frac{1}{1-t^2}}, $$ and so $$ L=\int_{-1}^{1}\sqrt{1+|h'(t)|^2}dt\ge\int_{-1}^1 \sqrt{\frac{1}{1-t^2}}dt=\pi. $$

A similar reasoning applies for any other point: Assume $p=(1,0,\dots,0)$, then you can assume that the other point $q$ is of the form $(a,s,0,\dots,0)$. The path $\gamma(t)=(t,\sqrt{1-t^2},0,\dots,0)$ has length $$ L=\int_{a}^{1}|\gamma'(t)|dt=\int_{a}^1 \sqrt{\frac{1}{1-t^2}}dt =-arccos(t)|_{a}^1=arccos(a)=arccos(\langle p,q\rangle). $$ By a similar argument as above, any other path has the form $(t,h(t))$ with $\sqrt{1+|h'(t)|^2}\ge \sqrt{\frac{1}{1-t^2}}$, so the result follows.