Writing rotations on $S^2$ as the product of two reflections on $S^2$. I am trying to prove that 
$\textbf{Prop:}$ The isometries of $S^2$ can each be written as the composition of at most three reflections.  
$\textbf{Work so far:}$
The identity can be written as the composition of no reflections. Reflections can be written as the composition of one reflection.
$\textbf{What I still need:}$
I need to show that rotations are the product of two reflections.  This would imply further that twist reflections are the product of 3 reflections, and I would be done.
$\textbf{Note:}$ I would appreciate more of a hint / some guidance here rather than a complete proof.  I am trying to prove this without showing / assuming that if $f,g$ are isometries in $S^2$ and $p_1,p_2,p_3$ are points not all lying on some great circle, then $f(p_i)=g(p_i)$ for $i=1,2,3$ implies $f=g$.
 A: (Oops, I just re-read your question, and your final observation is something you're trying not to assume. Well, I'll leave this up as a hint in case you do decide you want to use it.)
I would take a different approach than you: Let $p_1$, $p_2$, and $p_3$ be the points $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ respectively, and let $f$ be a given isometry. Show that you can find reflections $R_1$, $R_2$, and $R_3$ such that


*

*$R_1$ takes $p_1$ to $f(p_1)$

*$R_2$ leaves $f(p_1)$ fixed and takes $R_1(p_2)$ to $f(p_2)$, and

*$R_3$ leaves $f(p_1)$ and $f(p_2)$ both fixed and takes $R_2(R_1(p_3))$ to $f(p_3)$
Essentially you put each of the $p_i$ in the right place successively, making sure not to mess-up the previously placed points. Then use your final observation to show that $f = R_3 \circ R_2 \circ R_1$. I'd consider this a hint and not a worked-out proof as you still need to demonstrate that such reflections exist.
I'll confess that I'm not 100% sure that this will work, but it seems promising. I'd bet you could also use three other (non-co-circular) points but I think these particular ones will make it easier to show you can find the desired reflections as they are all at $90^{\circ}$ to each other.
