Let $\kappa$ be an uncountable, regular cardinal. We call $\kappa$ ineffable iff for every sequence $(A_\xi \colon \xi < \kappa)$ of subsets $A_\xi \subseteq \xi$ there is a stationary subset $S \subseteq \kappa$ such that for $\xi < \xi'$ in $S$: $A_\xi = A_{\xi'}\cap \xi$.
I want to show that $\kappa$ is weakly compact in the sense that it is inaccessible and does have the tree-property. I already proved that $\kappa$ is inaccessible, but now I feel a bit lost, which sequence to consider in order to verify the tree-property. Any hint would be highly appreciated.
Let $(T,<_T)$ be a $\kappa$-tree. Consider $$f \colon \kappa\rightarrow \kappa, \ \alpha \mapsto \left|\bigcup_{\beta < \alpha} T_\beta\right|$$ Then $f$ is monotone and continuous, so that there is a club $C \subseteq\kappa$ such that $f(\xi) \le \xi$ for every $\xi \in C$.
Now, we may recursively construct an isomorphic copy $T' \subseteq \kappa$ of $T$ such that $\bigcup_{\beta < \xi} T_\beta' \subseteq \xi$, whenever $\xi \in C$.
Consider the sequence $(A_\xi \colon \xi < \kappa)$, where $A_\xi$ is a chosen branch through $\bigcup_{\beta < \xi} T_\beta'$ whenever $\xi \in C$ and $A_\xi = \emptyset$ otherwise. As $\kappa$ is ineffable, we find a stationary subset $S \subseteq \kappa$ such that for $\xi < \xi'$ in $S$: $A_\xi = A_\xi' \cap \xi$. Then the $A_\xi$, $\xi \in S$ are initial segments of the cofinal path $A = \bigcup_{\xi \in S} A_\xi$ through $T'$, which witnesses a cofinal branch through $T$.