Given $F:[\omega_2]^{<\omega}\to[\omega_2]^{\aleph_0}$ as above, we first claim the existence of an ordinal $\omega_1\leq\alpha<\omega$ that is closed under $F$, i.e., $s\in [\alpha]^{<\omega}$ implies $F(s)\subseteq\alpha$. For this, let $\alpha$ be the limit of the sequence $\omega_1=\alpha_0<\alpha_1<\cdots$ where $\alpha_{n+1}$ is sufficiently large that $F(s)\subseteq \alpha_{n+1}$ for $s\in [\alpha_{n}]^{<\omega}$.

Given $\alpha$ as above, construct similarly the ordinal $\omega\leq\xi<\omega$ so that $f^{-1}_\alpha[\xi]$, that is, $\{\beta<\alpha:f_\alpha(\beta)<\xi\}$, is closed under $F$. This can be done similarly: let $\xi$ be the limit of the sequence $\omega=\xi_0<\xi_1<\cdots$ where $\xi_{n+1}$ is chosen so that if $s$ is a finite subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_n\}$, then $F(s)$ (which is a subset of $\alpha$) is a subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_{n+1})\}$. Now $X_{\alpha,\xi}$ is closed under $F$.

Given $F:[\omega_2]^{<\omega}\to[\omega_2]^{\aleph_0}$ as above, we first claim the existence of an ordinal $\omega_1\leq\alpha<\omega$ that is closed under $F$, i.e., $s\in [\alpha]^{<\omega}$ implies $F(s)\subseteq\alpha$. For this, let $\alpha$ be the limit of the sequence $\omega_1=\alpha_0<\alpha_1<\cdots$ where $\alpha_{n+1}$ is sufficiently large that $F(s)\subseteq \alpha_{n+1}$ for $s\in [\alpha_{n}]^{<\omega}$.

Given $\alpha$ as above, construct similarly the ordinal $\omega\leq\xi<\omega_1$ so that $f^{-1}_\alpha[\xi]$, that is, $\{\beta<\alpha:f_\alpha(\beta)<\xi\}$, is closed under $F$. This can be done similarly: let $\xi$ be the limit of the sequence $\omega=\xi_0<\xi_1<\cdots$ where $\xi_{n+1}$ is chosen so that if $s$ is a finite subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_n\}$, then $F(s)$ (which is a subset of $\alpha$) is a subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_{n+1})\}$. Now $X_{\alpha,\xi}$ is closed under $F$.