FALSE: every Borel subset of the real line is a countable union of countable intersections of open and closed sets

Every closed subset $F$ of $\mathbb{R}$ is a countable intersection of open sets: if $F\mathrm{\ne }\mathrm{\varnothing }$, use $O_j={\bigcup }_{a\in F}(a-1\mathrm{/}(j+1),a+1\mathrm{/}(j+1))$. Points of $F$ belong to every $O_j$; a point outside $F$ has an interval disjoint from $F$, and hence misses $O_j$ for sufficiently large $j$. For $F=\mathrm{\varnothing }$ use the constant empty sequence. Consequently any countable intersection of open and closed sets is a countable intersection of open sets: replace each closed factor by this explicit sequence, leave an open factor as its constant sequence, and enumerate pairs of indices by finite diagonals. Such a set is ${\mathrm{\Pi }}_2^0$, since its complement is a countable union of closed sets. Therefore every set in the claimed class is ${\mathrm{\Sigma }}_3^0$.

The map $j$ in L3 constructs a Cantor copy in the real line. Put $K=j[\mathcal{C}]$ with its subspace topology. The real line is Hausdorff: two distinct reals $s,t$ have disjoint open intervals of radius $\mathrm{\mid }s-t\mathrm{\mid }\mathrm{/}3$ around them. The subspace $K$ is therefore also Hausdorff, by intersecting these separating intervals with $K$. As a map onto $K$, $j$ is a continuous bijection: for open $O$ in $\mathbb{R}$, the preimage of $K\cap O$ is $j^{-1}[O]$. L3 gives compactness of its domain, so L4 gives a homeomorphism $\mathcal{C}\to K$. L4 also makes $K$ compact and closed in $\mathbb{R}$. Thus this explicit ternary-series map supplies the required homeomorphic Cantor subset without choice.

By L2 and step 1.2 the hypotheses of L1 hold for $\mathbb{R}$, so choose $D\in {\mathrm{\Pi }}_3^0(\mathbb{R})\setminus {\mathrm{\Sigma }}_3^0(\mathbb{R})$. This is a Borel set: the first three ranks are obtained from opens by complements and countable unions, operations preserving the Borel sigma-algebra. Suppose the statement were true. It would place $D$ in the displayed class and thus, by step 1.1, in ${\mathrm{\Sigma }}_3^0$, contradicting its choice. Hence the statement is false. The only choice input is L1; the conversion in step 1.1 makes no simultaneous choices of representations.