Strict Borel hierarchy in every uncountable Polish space

Apply F2 with A1 to X itself, obtaining a Cantor subspace. X is metrizable since it is Polish. Thus F1 with A1 gives $D\in {\mathrm{\Pi }}_{\alpha }^0(X)\setminus {\mathrm{\Sigma }}_{\alpha }^0(X)$.

For $\alpha >1$, each union of sets of lower $\mathrm{\Pi }$ rank allowed at rank $\alpha$ is also allowed at rank $\beta$. For $\alpha =1$, fix a compatible metric d: every open U is ${\bigcup }_n\{x:(\mathrm{\forall }y\mathrm{\notin }U)d(x,y)\ge 1\mathrm{/}(n+1)\}$, a union of closed sets. Closedness follows from the triangle inequality, and equality from the ball criterion for openness; if U=X the condition is vacuous, and if U is empty take y=x to exclude every x. Thus ${\mathrm{\Sigma }}_{\alpha }^0\subseteq {\mathrm{\Sigma }}_{\beta }^0$ also at rank one. The constant sequence D puts D in ${\mathrm{\Sigma }}_{\beta }^0$, proving this inclusion proper by step 1.1. Complementation gives ${\mathrm{\Pi }}_{\alpha }^0\subseteq {\mathrm{\Pi }}_{\beta }^0$ and its properness, since $X\setminus D$ belongs to the latter but not the former. QED.