Proving Sylow's first theorem Let $p$ be prime and $G$ a group such that $|G| = p^n k$, where $p \nmid k$. We want to show that $G$ has at least one Sylow $p$-subgroup.


  Let $\mathcal{S} := \{ S \subset G | \, |S| = p^n\}, |\mathcal{S}| := N$. Then, $N = \binom{p^n k}{p^n} \equiv k \text{ (mod } p)$. Let $G$ act on $\mathcal{S}$ by $(g, S) \mapsto gS = \{gs| \, s \in S\}$. Let $S_1, \dots, S_r$ represent the disjoint orbits of $\mathcal{S}$. Then, $p$ cannot be a divisor of the cardinality of all orbits, because $N \equiv k \text{ (mod } p)$. Choose $S_i$ such that $p \nmid |GS_i|$. Then, $\operatorname{Stab}_G(S_i) = S_i$ and thus $S_i$ is a Sylow $p$-subgroup.


Why is $\operatorname{Stab}_G(S_i) = S$? We defined the stabilizer as follows:

$$\operatorname{Stab}_G(x) := \{g \in G| \, gx = x\}$$

Maybe I'm just confusing things, but if we take $g \in G \setminus S_i$, isn't still $gS_i = S_i$?
You are confusing yourself by the notation. Also observe that the proof requires the notion of stabilizers of sets not points. But let's use your notation. Let $T \in S_i$, where the cardinality of orbit $S_i$ is not divisible by $p$. Put $H=Stab_G(T) = \{g \in G | gT=T\}$. We are going to argue that $H$ is a Sylow $p$-subgroup. First, since $p$ does not divide $|S_i| = [G:H]$, it follows that $p^n | |H|$, hence $p^n \leq |H|$. Since $H$ stabilizes $T$ by left multiplication, for any $t \in T$ we have $Ht \subseteq T$. Hence $|H| = |Ht| \leq |T| = p^n$. We conclude that $|H| = p^n$ and we are done.
I surmise that you took the proof in the wrong direction: we do not show that the stabiliser equality, but show that the stabiliser is a sub-group with the correct order.
I think also you meant to say that $p$ cannot divide the cardinality of the orbit of $S_i$, not the order of $S_i$, where $S_i$ is a subgroup in that orbit. Then let the orbit containing $S_i$ be $\mathscr O$.
So $|\mathscr O|=\mid G\mid/\mid H\mid$, where $H$ is the stabiliser of $S_i$.
Thus all the powers of $p$ appearing in the factorisation of $\mid G\mid$ must divide the order of $H$. This is equivalent with saying that $p\mid |H|$.
If the proof is fine in the direction described, tell me, for I cannot see how your proof is going to end. Regards.
Edit The above proof is not complete: we also have to show that the cardinality of $H$ is exactly $p^n$, not just divisible by $p^n$.
Since $H$ stabilises $S_i$, we know that, for every $s\in S_i$, $Hs\subset S$, and hence $\mid Hs\mid=\mid H\mid\le p^n$.