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}$$\frac{1}{|Gt|}$) for $s_i$ which yield distinct orbits. The sum on the right is equal to $1+…+1$ summed $|\{ Orbits\}|$ times since $\sum_{t \in Gs_i}\frac{1}{|Gt|}= 1$ by part one of the exercise in the text.
By the Orbit-Stabilizer Theorem, we have $\sum_{s \in S}\frac{1}{|Gs|}= \sum_{s \in S}\frac{|G_s|}{|G|} = \frac{1}{|G|}\sum_{s \in S}|G_s|$. Now it must be shown that $\sum_{s \in S}|G_s| = \sum_{x \in G}f(x)$. I proceed by demonstrating a bijection between the disjoint unions of each stabilizer and, for each $x\in G$, the set $F(x)$ which consists of all $s\in S$ such that $xs= s$.
Define a map from $\bigsqcup_{x \in G}F(x)$ to $\bigsqcup_{s \in S}G_s$ by $f_x(s)\mapsto g_s(x)$ where each $f_x: F(x)\rightarrow\bigsqcup_{x \in G}$ is injective and partition $\bigsqcup_{x \in G}$ (define $g_s(x)$ similarly.) This map is clearly bijective and we are done.
Is this proof correct? It seems much more complicated than simply considering the pairs $(g,s)$, but I thought I might as well ask.
It's basically fine, and it's actually much more similar to the standard proof than you probably realize. Let's examine what you mean by the disjoint unions. You treat $\bigsqcup_{x\in G}$ like a set, but that's nonsensical. What you really mean is for example $$ 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 $S^x$.