Hi. Here's the proof I've learned. It does not separate the cases $\alpha=2$ and $\alpha>2$. I hope I am not misunderstanding your question.
Theorem: Let $X$ be an uncountable Polish space. Then for any $\alpha<\omega_1$, there is a set $A\in \Sigma^0_{\alpha} \backslash \Pi^0_{\alpha}$. There is also for each $\alpha<1$ a set in $\Delta^0_{\alpha} \backslash \bigcup_{\beta<\alpha} \Sigma^0_{\beta} $
Proof: Let $X$ be an uncountable Polish space. The theorem is quick in the case $\alpha=1$ since if every open set was also closed then every point and every subset of $X$ is open which gives $X$ the discrete topology. But $X$ was Polish. Let $\alpha>1$. Recall that is $X$ is uncountable Polish then there is an embedding of $2^{\omega}$ into $X$. So there is a subspace $C \subseteq X$ homeomorphic to the Cantor space. $C$ has to be closed in $X$. Since having a universal set implies non-self-duality and by the existence of universal sets, there is a set $A\subseteq C$ which is $ \Sigma^0_{\alpha} $ but not $ \Pi^0_{\alpha} $ in the subspace topology on $C$. So in $X$, $A$ is the intersection of $ \Sigma^0_{\alpha} $ set and a closed set so it is $ \Sigma^0_{\alpha} $. If $A$ were also $\Pi^0_{\alpha} $ then $A=A \cap C$ would also be $ \Pi^0_{\alpha} $ in the subspace topology on $C$ which is a contradiction. For the second part, let $2\leq \alpha$. We show there is an $A \subseteq 2^{\omega}$ which is in $\Delta^0_{\alpha}$ but not in any $\Sigma^0_{\beta}$ for any $\beta<\alpha$. If $\alpha$ is a successor, say $\alpha=\beta+1$ then let $A\subseteq 2^{\omega}$ in $\Sigma^0_{\beta} \backslash \Pi^0_{\beta}$ and let $B \subseteq 2^{\omega}, B\in \Pi^0_{\beta} \backslash \Sigma^0_{\beta} $ . Define $C\subseteq 2^{\omega}$ such that we have $C(x) \leftrightarrow (x(0)=0 \wedge x(i+1)\in A \forall i) \vee (x(0)=1 \wedge x(i+1)\in B \forall i$. Then $C \in \Delta^0_{\alpha}$