The exercise does not have a solution included, so I'd be thankful, if people here can verify my solution and/or fill any gaps I might have. Also it'd be great if someone could point out a way to shorten the proof.
So here goes my interpretation of the statement I am trying to prove:
Let $A$ be any non-empty set. Let $G$ be a subgroup of $S(A)$, ie. group of permutations on $G$
Let $O: A \rightarrow P(A)$ such that $O(u) = \{g(u) : g \in G\}$. We will call it the orbit of u. Hence $O(u) \subseteq A$.
Let $C: A \rightarrow P(G)$ such that $C(u) = \{g \in G : g(u) = u \}$. We will call it the stabilizer of u. Hence $C(u) \subseteq G$.
We want to prove that $|O(u)|=|G|/|C(u)|$.
Since $|G|/|C(u)|$ is the number of cosets of $C(u)$, then it will suffice to prove that there is a onto function from $O(u)$ to the set of cosets of $C(u)$ in G.
Which can be written as $\exists f$ such that $f_{u}: O(u) \rightarrow G \backslash C(u)$ and is onto.
Proof:
Let $f_{u}: O(u) \rightarrow G \backslash C(u)$ and let $f_{u}(k) = gC(u) : g(u) = k$
We want to look at set $\{gC(u) : g(u) = k\}$ and prove that it has exactly one element.
Suppose we have $g,n : g(u) = k,n(u) = k$, we need to prove that $gC(u) = nC(u)$.
Since we're talking about cosets, it will suffice to prove that $g^{-1}n \in C(u)$. Which is equivalent to $g^{-1}n(u) = u$.
Now $n(u)=k$ and $g^{-1}(k)=u$ since $g(u)=k$. Hence $g^{-1}n(u) = u$ and $gC(u) = nC(u)$.
Now we know that $f_{u}$ is well defined.
Now we need to verify that it's injective and surjective.
Suppose that $f_{u}(k)=f_{u}(t) : k \neq t$.
Then $gC(u) : g(u)=k$ and $pC(u) : p(u)=t$.
If $gC(u)=pC(u)$ then $g^{-1}p \in C(u)$ $<=>$ $g^{-1}p(u) = u$.
But $p(u)=t$ and $g^{-1}(t) \neq u$ because $g^{-1}(k) = u$.
Hence we have a contradiction and $f_{u}$ is injective.
Now to check the surjectivity part.
Lets take a sample coset of $C(u)$, $gC(u)$, we know that g(u) is well-defined. Let g(u)=k. Then $f_{u}(k)=gC(u)$ by definition.
Hence $f_{u}$ is surjective.