Why is set required to be finite in the Orbit-Stabilizer Theorem In the Orbit Stabilizer Theoren (found here) we have a group acting on a finite set $X$. I don't see in the proof given how the finiteness of the set is used. 
I get that if one wants to write $\lvert G\rvert / \lvert Stab_G(x)\rvert$, then we need something to be finite to form the fraction, but in general I don't see it.
 A: You're right, it's not necessary for the statement that $|\operatorname{Orb}(x)|=[G:\operatorname{Stab}_G(x)]$. $\newcommand\Orb{\operatorname{Orb}}\newcommand\Stab{\operatorname{Stab}}$
After all, the proof is as follows. Let $H=\Stab_G(x)$. We define a bijection $\phi : G/\Stab_G(x)\to \Orb(x)$ by $\phi(gH)=gx$. This is a well defined function since if $g_1H=g_2H$, $g_2^{-1}g_1H=H$, so $g_2^{-1}g_1\in H$. Thus $g_2^{-1}g_1x=x$, and $g_1x=g_2x$.
The proof of injectivity is the proof of well-definedness in reverse. If $\phi(g_1H)=\phi(g_2H)$, then $g_1x=g_2x$. Thus $g_2^{-1}g_1x=x$, so $g_2^{-1}g_1\in H$, and thus $g_1H=g_2H$.
Finally, for surjectivity, we observe that if $gx\in \Orb(x)$, then $\phi(gH)=gx$. Thus $\phi$ is surjective.
Side note
In fact, we can prove the stronger result that as left $G$-sets (assuming a left $G$-action to start with). $\Orb(x) \cong G/\Stab_G(x)$, where $G/\Stab_G(x)$ denotes the left coset space of $\Stab_G(x)$. The only additional thing we need to prove is that the bijection $\phi$ we constructed above is $G$-equivariant, and this follows from the fact that
$$\phi(g_1g_2H)=g_1g_2x=g_1(g_2x)=g_1\phi(g_2H).$$
