Proof
(related to Proposition: Transitive Recursion)
 By hypothesis, $X$ is a set and $\Omega$ is the proper class of all ordinals and $\omega\in\Omega$ is a given ordinal number.
 It is sufficient to show that the function $f:\mathbb \Omega\to X$ defined by specifying the value of $f(\alpha)$ for all $\alpha\subseteq \omega$ and for all $\beta\supset\omega$ with a recursion formula $f(\beta)=\mathcal R(f(\alpha)\mid \alpha \subset\beta)$ is welldefined (i.e. unique) for all ordinal numbers.
 Assume, $f$ defined like this is not welldefined for some ordinal numbers.
 Since ordinal numbers are wellordered, there is the smallest ordinal number $\omega_0\in\mathbb \Omega$ for which $f$ is not welldefined.
 But then for $\alpha \subset \omega_0$ the values $f(\alpha)$ are welldefined and $f(\omega_0)=\mathcal R(f(\alpha)\mid \alpha \subset \omega_0).$
 This is a contradiction.
 Therefore, $f$ is welldefined for all ordinal numbers.
∎
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References
Bibliography
 Toenniessen, Fridtjof: "Topologie", Springer, 2017