Prove a set is well-ordered

Let's first prove this prelimiar lemma: if $\kappa$ and $\lambda$ are well-ordered sets, then $\kappa \times \lambda$ with the antilexicographic order is well-ordered.

Proof: Let $S \subseteq \kappa \times \lambda$, and let $S_2 = \{ \alpha \mid \exists \beta \colon (\beta, \alpha) \in S \}$. Since $S_2 \subseteq \lambda$, it has a minimum element, say $\alpha_0$. Now take $S_1 = \{ \beta \mid (\beta, \alpha_0) \in S\}$. Since $S_1 \subseteq \kappa$, it also has a minimum element, say $\beta_0$. It follows that $(\beta_0, \alpha_0)$ is the minimum of $S$, and so $\kappa \times \lambda$ is well-ordered.

We can now show that $FS(\alpha, \beta)$ is well-ordered by transfinite induction on $\beta$.

If $\beta = 0$, then $FS(\alpha, 0) = 1$ contains a single function, the empty function, and so it is well-ordered.

If $\beta = \gamma + 1$, then we have \[ FS(\alpha, \beta) = FS(\alpha, \gamma) \times \alpha, \] the bijection being $f \mapsto (f|_\gamma, f(\gamma))$. By inductive hypothesis $FS(\alpha, \gamma)$ is well-ordered, since $\alpha$ is an ordinal it is also well-ordered, and so it is easy to check that the product is well-ordered.

If $\beta = \sup_{\gamma < \beta} \gamma$, then \[ FS(\alpha, \beta) = \bigcup_{\gamma < \beta} FS(\alpha, \gamma), \] as every finitely supported function $f \colon \beta \rightarrow \alpha$ is in fact supported on $\gamma + 1$, where $\gamma = \max \{\delta < \beta \mid f(\delta) \neq 0\}$, and the maximum exists because the set is finite. But the union of well-ordered sets that are initial segments of each other is well-ordered (it's a union of ordinals), so $FS(\alpha, \beta)$ is well ordered.