Question on a proof of Kuratowski's Theorem in Kechris's Classical Descriptive Set Theory Currently I am reading Kechris's Classical Descriptive Set Theory.

Theorem $22.18$: (Kuratowski) Let $(X,\tau)$ be a Polish space and $A_n\subseteq X$ be $\Delta_\xi^0(X,\tau).$
  Then there is a Polish topology $\tau'\supseteq \tau$ such that $\tau'\subseteq\Sigma_\xi^0(X,\tau)$ and $A_n\in\Delta_1^0(X,\tau')$ for all $n.$

The following is a proof given by the author.

It is enough to prove this for a single set $A\in\Delta_\xi^0(X,\tau).$
  The proof is by induction on $\xi\geq 1.$
  For $\xi=1,$ take $\tau'=\tau.$
  For $\xi=2,$ both $A$ and $\sim A$ are $G_\delta,$ so Polish in the relative-$\tau$ topology. 
  Put on $X$ the direct sum $\tau'$ of these relative topologies. 
  So $U\in\tau'$ if and only if $U\cap A, U\setminus A$ are open in $A$ and $\sim A$ respectively. 
  This is clearly Polish, and $A$ is $\Delta_1^0$ in $\tau'.$
  Also, $\tau'\subseteq \Delta_2^0(X,\tau)\subseteq \Sigma_2^0(X,\tau).$
Let now $\xi$ be a limit ordinal. 
  Then $A=\bigcup_n A_n = \bigcap_n B_n$ with $A_n,B_n\in \Delta_{\xi_n}^0(X,\tau),\xi_n<\xi.$
  Let $\tau_n',\tau_n{''}$ be topologies that work for $A_n,B_n$ respectively. 
  Let $\tau'$ be the topology generated by $\bigcup_n (\tau'_n\cup \tau''_n).$
  Then it is Polish and clearly $A\in\Delta_1^0(X,\tau').$
  Since every set in $\tau'\cup\tau''$ is in $\Sigma_\xi^0(X,\tau),$ clearly $\tau'\subseteq\Sigma_{\xi}^0.$
Finally, let $\xi = \eta+1\geq 3$ be successor. 
  Then $A=\lim_nA_n,A_n\in\Delta_\eta^0(X,\tau).$
  Let $\tau^*\supseteq\tau$ be Polish with $\tau^*\subseteq\Sigma_\eta^0(X,\tau)$ and $A_n\in\Delta_1^0(X,\tau^*)$ for all $n.$
  Then $A\in\Delta_2^0(X,\tau^*).$
  Apply now the case $\xi=2$ to $(X,\tau^*)$ to obtain $\tau'\supseteq\tau^*$ with $A\in\Delta_1^0(X,\tau')$ and $\tau'\subseteq\Sigma_2^0(X,\tau')\subseteq\Sigma_{\eta+1}^0(X,\tau)=\Sigma_\xi^0(X,\tau).$

The following are my questions on the proof.

($1$) In first paragraph, how to show that $\tau'$ is Polish and $\tau'\subseteq\Delta_2^0(X,\tau)$?
$(2)$ In second paragraph, how to obtain that $A_n,B_n\in\Delta_{\xi_n}^0(X,\tau)?$ 
  I can only get $A_n\in\Pi_{\xi_n}^0(X,\tau)$ and $B_n\in\Sigma_{\xi_n}^0(X,\tau).$
($3$) In second paragraph, why is it $A\in\Delta_1^0(X,\tau')?$

 A: Below when I cite a theorem a proposition or a page I am referring to the Kechris's book you cited in your question.


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*By thm 3.11 $A$ and $\sim A$ are Polish spaces since $A$ and $\sim A $ are $G_{\delta}$ sets. 
So there are $ (A,d)$ and $(\sim A, d')$  metric spaces such that the topology induced by $d$ and $d'$ are the subspace topology of $ A$ and $ \sim A$. It is straightforward to verify, using $d$ and $d'$ that the sum topology $\tau'$ is such that $ (X,\tau')$ is Polish (this is stated already in proposition 3.3 without a proof, but see the comments on p. 3 regarding the sum of metric spaces). 

*By the definition on p. 68 there are sequences $\langle \alpha_{n},\beta_{n},A_{n},B_{n} \ | \ n \in \omega \rangle$ such that  $\alpha_{n} \in \xi $ and $\beta_{n} \in \xi $,  $ A = \bigcup_{n\in \omega} A_{n} = \bigcap_{n\in\omega}B_{n} $ and for each $n \in \omega$ we have  $A_{n} \in \Pi^{0}_{\alpha_{n}}$ and $B_{n} \in \Sigma^{0}_{\beta_{n}}$. Define $\xi_{n} = (\max\{\alpha_{n},\beta_{n}\} + 1) < \xi$, then $ \Sigma^{0}_{\alpha_{n}} \subseteq \Delta^{0}_{\xi_{n}} $ and $\Pi^{0}_{\beta_{n}} \subseteq \Delta^{0}_{\xi_{n}}$. 

*We have $\Delta^{0}_{1}(X,\tau') =\{ W \subseteq X \ | \ W \ \text{is a clopen in} \ \tau' \}$. We only need to verify that $ A$ is a clopen. We have $ A \cap A $ is open in $A$ and $ A \setminus A $ is open in $\sim A$, thus $ A$ is open in $(X,\tau') $. $ \sim A \cap A $ is open in $ A $ and $ \sim A \setminus \sim A $ is open in $\sim A $, so $ A $ is closed in $(X,\tau')$.
