Why use transfinite induction to prove? I think the inclusion relation is trivial by transfinite recursion.
$\mathbf{11.B}$ The Borel Hierarchy
Assume now that $X$ is metrizable, so that every closed set is a $G_\delta$ set. Let $\omega_1$ be the first uncountable ordinal, and for $1 \leq \xi < \omega_1$ define by transfinite recursion the classes $\sum_\xi^0 (X)$, $\prod_\xi^0 (X)$ of subsets of $X$ as follows: \begin{align} \sum_1^0 (X) &= \{U \subseteq X:U \text{ is open}\}, \\ \prod_\xi^0 (X) &= {} \sim \sum_\xi^0 (X), \\ \sum_\xi^0 (X) &= \left\{ \bigcup_n A_n : A_n \in \prod_{\xi_n}^0 (X), \xi_n < \xi, n \in \mathbb{N}, \text{ if } \xi >1\right\}. \end{align} In addition let $$\Delta_\xi^0 (X) = \sum_\xi^0 (X) \cap \prod_\xi^0 (X)$$ be the so-called ambiguous classes.
Traditionally, one denotes by $G(X)$ the class of open subsets of $X$, and by $F(X)$ the class of closed subsets of $X$. For any collection $\mathcal E$ of subsets of a set $X$, let \begin{align} \mathcal E_\sigma &= \left\{ \bigcup_n A_n : A_n \in \mathcal E, n \in \mathbb N \right\}, \\ \mathcal E_\delta &= \left\{ \bigcap_n A_n : A_n \in \mathcal E, n \in \mathbb N \right\}. \end{align} Then we have $\sum_1^0 (X) = G(X)$, $\prod_1^0 (X) = F(X)$, $\sum_2^0 (X) = (F(X))_\sigma = F_\sigma (X)$, $\prod_2^0 (X) = (G(X))_\delta = G_\delta (X)$, $\sum_3^0 (X) = (G_\delta(X))_\sigma = G_{\delta\sigma} (X)$, $\prod_3^0 (X) = (F_\sigma(X))_\delta = F_{\sigma\delta} (X)$, etc. (Also, $\Delta_1^0 (X) = \{ A \subseteq X: A \text{ is clopen}\}$.) In general, an easy $\color{red}{\text{transfinite induction}}$ shows that $$\sum_\xi^0 (X) \cup \prod_\xi^0 (X) \subseteq \Delta_{\xi+1}^0 (X),$$
Thank you in advance.