Alternative proof of Burnside’s Lemma Burnside’s Lemma states:

Let $G$ be a finite group acting on a finite set $S$. For each $x$ $\in$ $G$ define $f(x)$ = number of elements $s$ $\in$ $S$ such that $xs$ = $s$. Prove that the number of orbits of $G$ in $S$ is equal to $\frac{1}{|G|}\sum_{x \in G}$ $f(x)$.

This is typically solved by considering the subset $\{(g,s) : gs = s\}$ of $G\times S$, however I used a different approach. Here is the proof I gave:
The number of orbits of $G$ in $S$ equals $\sum_{s \in S}$ $\frac{1}{|Gs|}$ where $Gs$ is the orbit of $s$ in $G$. This is because $\sum_{s \in S}$ $\frac{1}{|Gs|}$ = $\sum_{s_i}$($\sum_{t \in Gs_i}$$\frac{1}{|Gt|}$) for $s_i$ which yield distinct orbits. The sum on the right is equal to $1+…+1$ summed $|\{ Orbits\}|$ times since $\sum_{t \in Gs_i}\frac{1}{|Gt|}= 1$ by part one of the exercise in the text.
By the Orbit-Stabilizer Theorem, we have $\sum_{s \in S}\frac{1}{|Gs|}= \sum_{s \in S}\frac{|G_s|}{|G|} = \frac{1}{|G|}\sum_{s \in S}|G_s|$. Now it must be shown that $\sum_{s \in S}|G_s| = \sum_{x \in G}f(x)$. I proceed by demonstrating a bijection between the disjoint unions of each stabilizer and, for each $x\in G$, the set $F(x)$ which consists of all $s\in S$ such that $xs= s$.
Define a map from $\bigsqcup_{x \in G}F(x)$ to $\bigsqcup_{s \in S}G_s$ by $f_x(s)\mapsto g_s(x)$ where each $f_x: F(x)\rightarrow\bigsqcup_{x \in G}$ is injective and partition $\bigsqcup_{x \in G}$ (define $g_s(x)$ similarly.) This map is clearly bijective and we are done.
Is this proof correct? It seems much more complicated than simply considering the pairs $(g,s)$, but I thought I might as well ask.
 A: It's basically fine, and it's actually much more similar to the standard proof than you probably realize. Let's examine what you mean by the disjoint unions. You treat $\bigsqcup_{x\in G}$ like a set, but that's nonsensical. What you really mean is for example
$$
f_x:F(x)\to \bigsqcup_{y\in G}F(y)
$$
(or you can use $x$'s on the RHS too; doesn't matter). So what does $\bigsqcup_{x\in G}F(x)$ mean? The natural construction is
$$
\bigsqcup_{x\in G}F(x)
= \bigcup_{x\in G} \{x\}\times F(x)
= \bigcup_{x\in G} \left\{(x, s) \mid s\in F(x)\right\}.
$$
Simlarly,
$$
\bigsqcup_{s\in S}G_s 
= \bigcup_{s\in S} \{s\}\times G_s
= \bigcup_{s\in S} \left\{(s, x) \mid x\in G_s\right\}.
$$
With this, the bijection simply becomes
$$
(x,s) = f_x(s) \mapsto g_s(x) = (s,x),
$$
so in the end, it's all equivalent to considering the set $\{(x,s)\mid xs=s\}\subset G\times S$.
My one issue with your proof is that you just write: "This map is clearly bijective and we are done." Is that really clear? It's clear that we basically just swap $x$ and $s$, so it sure looks like a bijection. But crucially, we need to realize that the sets are actually just "mirrors" of each other. That's only clear once we think about what defines $F(x)$ and $G_s$, and perhaps that was clear to you, but it should be mentioned in the proof.
In map theoretic terms, you could finish like this. Let $\phi$ be the purported bijective map, and $\psi:g_s(x)\mapsto f_x(s)$ the purported inverse. Then you must show that
$$
\phi(f_x(s)) \in \bigsqcup_{t\in S}G_t
\quad\textrm{and} \quad
\psi(g_s(x)) \in \bigsqcup_{y\in G}F(y).
$$
After a moment of thought, this is equivalent to showing that $s\in F(x) \iff x\in G_s$, which of course is immediate from definitions. It's then immediate as well that $\phi$ and $\psi$ are inverses, so $\phi$ is bijective. But without showing this, there is no guarantee that $\phi$ and $\psi$ have the right codomains, which would ruin the proof if false.
By the way, $F(x)$ is commonly written as $S^x$.
