Let $M_\alpha$ for $\alpha\in ON$ be transitive sets and let $M=\bigcup_{\alpha\in ON}M_\alpha$. Suppose that (i) for every $\alpha<\beta$, we have $M_\alpha\in M_\beta$ and (ii) for every limit $\lambda$, $M_\lambda=\bigcup_{\alpha<\lambda} M_\alpha$. Prove that if for arbitrarily large $\alpha\in ON$, $M_\alpha$ satisfies the comprehension schema, then $M$ satisfies the Comprehension schema.
Any help would be appreciated
The way I see it, you have to combine two observations, but actually get away with weaker assumptions:
Proposition. Let $(M_{\alpha} \mid \alpha \in \operatorname{On})$ be a sequence such that $M_{\alpha}$ is transitive, $M_{\alpha} \in M_{\beta}$ for all $\alpha < \beta \in \operatorname{On}$, $M_{\lambda} = \bigcup \{ M_{\alpha} \mid \alpha < \lambda \}$ for limit ordinals $\lambda$ and such that the set of $\alpha$'s such that $M_{\alpha}$ satisfies comprehension for $\Sigma_{1}$ formulas is unbounded. (We can get away with comprehension for $\Sigma_{0}$ formulas, if we know that our $M_{\alpha}$'s are closed under pairing.) Then $M := \bigcup \{ M_{\alpha} \mid \alpha \in \operatorname{On} \}$ satisfies full comprehension.
Proof. Let $\phi(v_{0}, \ldots, v_{n-1})$ be a formula in the language of set theory with exactly $v_{0}, \ldots, v_{n-1}$ as free variables and let $p_{1}, \ldots, p_{n-1} \in M$. We want to see that
$$\{x \in p_{1} \mid M \models \phi(x,p_{1}, \ldots, p_{n-1} \} \in M.$$
Fix $\alpha < \operatorname{On}$ such that $\{p_{1}, \ldots, p_{n-1} \} \subseteq M_{\alpha}$ and such that $\phi(v_{0}, p_{1}, \ldots, p_{n-1})$ is absolute between $M_{\alpha}$ and $M$. (Such an $\alpha$ exists by the Reflection Principle, because $(M_{\alpha} \mid \alpha \in \operatorname{On})$ is a strictly increasing and continuous sequence of transitive sets.) Let $\alpha < \beta \in \operatorname{On}$ be such that $M_{\beta}$ satisfies comprehension for $\Sigma_{0}$ formulas. Now, in $M_{\beta}$, the statement $$x \in p_{1} \wedge M_{\alpha} \models \phi(x,p_{1}, \ldots, p_{n})$$ can be expressed as a $\Sigma_{1}$ formula in paramaters $p_{1}, \ldots, p_{n}, M_{\alpha}$. (Again, we actually can do a little better. This formula can be set up in a way that it actually does not depend on the formula $\phi$ itself, but only on its rank in the Levy hierarchy.) Since it may also be expressed in a $\Pi_{1}$ way, we get (by the absoluteness of $\Delta_{1}$ formulas between transitive structures, that $$\{x \in p_{1} \mid M_{\alpha} \models \phi(x,p_{1}, \ldots, p_{n-1}) \} = \{x \in M_{\beta} \mid M_{\beta} \models \left( x \in p_{1} \wedge M_{\alpha} \models \phi(x, p_{1}, \ldots, p_{n-1}) \right)\} \in M_{\beta} \subseteq M.$$ qed