Blass's paired finite-modification classes form a Russell set

For distinct $r,s<\omega$ and either choices of complement, the corresponding Cohen reals have infinite symmetric difference. Indeed, beyond any prescribed finite set of bits, every condition has an extension assigning one fresh bit at coordinates $r$ and $s$ so that the two chosen versions disagree. Genericity meets each of these dense sets. Also $a_r\mathrm{△}(\omega \setminus a_r)=\omega$. Thus the finite-modification classes in different displayed positions are distinct; equivalence classes are either equal or disjoint. Each $f(n)$ therefore has exactly two elements and the family $R$ is pairwise disjoint.

Suppose for contradiction that $c$ is a choice function on $\{f(k):k\in K\}$ for an infinite $K\subseteq \omega$ in $N$. By F1, $c$ is uniquely defined in $M[G]$ from $f$, ordinals, and finitely many real parameters $s_1,\dots ,s_t$ from $S\setminus \{f\}$. For each $s_i$, fix one ground finite set $z_i$, one coordinate $m_i$, and one sign such that $s_i$ is $a_{m_i}\mathrm{△}{z}_i$ or $(\omega \setminus a_{m_i})\mathrm{△}{z}_i$. Since $K$ is infinite and $\{m_1,\dots ,m_t\}$ is finite, the least element $k$ of their difference exists without Choice.

The value $c(f(k))$ is one of the two classes in $f(k)$; interchange the labels if necessary and suppose it is $\delta (a_k)$. By F4, some finite $p\in G$ forces both the unique defining formula for $c$ and this value assertion. Choose $b$ above every $j$ with $(k,j)\in \mathrm{dom}(p)$, taking $b=0$ if there is none. Flip precisely the bits $(k,j)$ for $j\ge b$. By F2 the induced automorphism fixes $p$, every ordinal, and each named $s_i$, while it interchanges $\delta (a_k)$ and $\delta (\omega \setminus a_k)$. It fixes $f$ because it merely swaps the two members of $f(k)$ and fixes every other value.

Apply F3 to the formula forced by $p$. Since the condition and every defining parameter are fixed, the same $p$ forces that the same uniquely defined function $c$ takes $f(k)$ to $\delta (\omega \setminus a_k)$. As $p\in G$, both value statements hold in $M[G]$. Step 1.1 says the two values are distinct, contradicting that $c$ is a function. The case in which the original value is $\delta (\omega \setminus a_k)$ is identical because the flip is an involution.

Hence no infinite subfamily of $R$ has a choice function. The function $f:\omega \to R$ already belongs to $N$, so $R$ is countably indexed there; step 1.1 supplies disjoint two-element pieces. By the source definition, $\bigcup R$ is therefore a Russell set. No family of representatives of the finite-modification classes was selected in $N$.