Every uncountable Solovay-model set of reals has a perfect subset

Use F2 to choose $\xi <\kappa$ such that $N=V[G_{\xi }]$ contains F1's real parameter; keep the finite ordinal parameters explicit. Fix an ambient enumeration $e:\omega \twoheadrightarrow {\mathbb{R}}^N$. If $A\subseteq N$, choose one $a_0\in A$ (available because $A$ is uncountable) and define $e_A(n)=e(n)$ when $e(n)\in A$, and $e_A(n)=a_0$ otherwise. This is an ambient omega-surjection onto $A$ with values in $M$, so F3 puts it in $M$, contradicting internal uncountability. Hence some $x\in A\setminus N$ exists.

Use F2 again to choose $\eta$ with $\xi <\eta <\kappa$ and $x\in V[G_{\eta }]$. Let $Q$ be the finite-condition collapse on $[\xi ,\eta )\times \omega$ and $H=G_{\eta }\upharpoonright ([\xi ,\eta )\times \omega )$. The disjoint-coordinate projection in F2 gives $V[G_{\eta }]=N[H]$, with $Q\in N$ and $Q$ small. By the defining valuation formula for $N[H]$ in F2, choose a $Q$-name $\dot{\tau }\in N$ with ${\dot{\tau }}_H=x$.

In $N$ form the downward-open set $D=\{q\in Q:(\mathrm{\exists }y\in {\mathbb{R}}^N)q{⊩}_Q\dot{\tau }=\check{y}\}$. The actual generic $H$ misses $D$, since the truth lemma would otherwise put $x={\dot{\tau }}_H$ in $N$. The set $D\cup \{q:q\perp D\}$ is dense and belongs to $N$, so genericity gives $p_0\in H$ incompatible with every member of $D$. Hence no extension of $p_0$ forces $\dot{\tau }$ equal to a real of $N$. Separately, the truth lemma and tail homogeneity give $p_1\in H$ forcing the fixed real-membership formula defining $A$ (with its localized real and ordinal parameters). Directedness of $H$ gives $p\in H$ below $p_0,p_1$. This $p$ has the exact forcing-newness premise of F5 and forces every branch interpretation to satisfy the definition of $A$; no external class formula “$\dot{\tau }\mathrm{\notin }N$” has been used in the forcing language.

Apply F5 below $p$. Every branch interpretation satisfies the same definition, so its compact injective image $P$ lies in $A$. The construction has a real code; since $M$ has all reals, that code lies in $M$, and F6 says internally that $P$ is nonempty perfect.