This seems like one these situations where one has to strengthen the claim in order to get through the induction. It may be much too complicated what I am doing here, but as a wise man once said, if it works it ain't stupid.
Let $\Sigma$ be the following set of $\{\in, \dot A\}$-formulas:
- all atomic formulas are in $\Sigma$
- $\Sigma$ is closed under $\wedge$ and $\neg$
- if $\varphi(x)$ is a $\in$-formula, then $\exists x\ \varphi(x)$ is in $\Sigma$
- if $\varphi(x, y)$ is in $\Sigma$ then $\{x\in \bigcup\dot A\mid \varphi(x, y)\}\in \dot A$ is in $\Sigma$
Essentially we allow $\dot A$ as long as we do not quantify over it.
I claim that the lemma holds true for all $\Sigma$-formulas.
For all $x$, $\gamma_1<\dots<\gamma_n<\alpha$ and $\varphi\in\Sigma$ we have:
$$(M_\alpha, \in, U_\alpha)\models\varphi(i_{0, \alpha}(x), \kappa_{\gamma_1},\dots, \kappa_{\gamma_n})$$
iff
$$(M, \in, U)\models\{\{\xi_1, \dots, \xi_n\}\in[\kappa]^{n}\mid \varphi(x, \xi_1,\dots, \xi_n)\}\in U^n$$
Annoyingly, we have to backtrack quite a bit to do it. First let me show that $(M, \in, U)$ satisfies a form of replacement with respect to $\Sigma$-formulas, that is for any $\Sigma$-formula $\varphi$ and parameter $p\in M$, the function $F:\kappa\rightarrow \mathcal{P}(\kappa)$ given by
$$F(\alpha)=\{\beta<\kappa\mid\varphi(\alpha,\beta, p)^M\}$$
is in $M$. We prove this by induction on the complexity of $\varphi$. The only nontrivial case arises when
$$\varphi(x, y, z)= \{u\in\bigcup \dot A\mid \psi(u, x, y, z)\}\in\dot A$$
Let $h:\kappa\rightarrow\kappa\times\kappa$ be a bijection in $M$. By induction, the function $G:\kappa\rightarrow\mathcal{P}(\kappa)$ given by
$$G(\delta)=\{\gamma<\kappa\mid\psi(\gamma, h(\delta)_0, h(\delta)_1, p)^M\}$$
is in $M$. By weak amenability, we have
$$X=\{\delta<\kappa\mid G(\delta)\in U\}\in M$$
But now
$$F(\alpha)=\{\beta<\kappa\mid h^{-1}(\alpha,\beta)\in X\}$$
and thus $F\in M$.
It follows immediately (from weak amenability) that $M$ satisfies the corresponding form of separation with respect to $\Sigma$-formulas that is:
For any $\Sigma$-formula $\varphi$ and parameter $p\in M$ we have
$$\{\alpha<\kappa\mid\varphi(\alpha, p)\}^M\in M$$
We now prove Los's theorem for the $\Sigma$-formulas in our context. Again, almost everyhing is trivial or works as in the usual proof. Note that the $\exists$-case goes through as in that case $\dot A$ is not allowed to appear in the formula. For the new part, suppose $\varphi=\{x\in\bigcup\dot A\mid \psi(x, y)\}\in \dot A$. We look at when
$$(M_1,\in, U_1)\models \{\beta< \kappa_1\mid\psi(\beta, [f])\}\in U_1$$
holds.
Note that $\{\beta< \kappa_1\mid\psi(\beta, [f])\}^{M_1}$ is really a set in $M_1$ and furthermore if we define $g:\kappa\rightarrow M$ by
$$g(\alpha)=\{\beta<\kappa\mid\psi(\beta, f(\alpha))\}^M$$
then $g\in M$ and
$$[g]=\{\beta< \kappa_1\mid\psi(\beta, [f])\}^{M_1}$$
Thus by definition of $U_1$,
$$(M_1,\in, U_1)\models [g]\in U_1\Leftrightarrow (M, \in, U)\models\{\alpha<\kappa\mid g(\alpha)\in U\}\in U$$ and thus
$$(M_1, \in, U_1)\models\varphi([f])\Leftrightarrow (M,\in, U)\models\{\alpha<\kappa\mid\varphi(f(\alpha))\}\in U$$
which is what we had to show.
Finally, we can tackle the lemma in our strenghtened form. Let us go through it to see that now, everything works out fine. Again the case $n=0$ is immediate, as our Los theorem for $\Sigma$ implies that the embeddings $i_{\alpha, \beta}$ are elementerary w.r.t. $\Sigma$-formulas. In the inductive step from $n-1$ to $n$:
$$(M_\alpha, \in, U_\alpha) \models \varphi(i_{0\alpha}(x), \kappa_{\gamma_1} ,\dots,\kappa_{\gamma_n})$$
$$\text{ iff } (M_{\gamma_n+1}, \in, U_{\gamma_{n+1}})\models \varphi(i_{0,\gamma_n+1}(x), \kappa_{\gamma_1} ,\dots,\kappa_{\gamma_n})$$
$$\text{ iff }(\ast)_0 (M_{\gamma_n}, \in, U_{\gamma_n}) \models \{\xi \lt \cup U_{\gamma_n} \mid \varphi(i_{0\gamma_n}(x), \kappa_{\gamma_1} ,\dots,\kappa_{\gamma_{n−1}}, \xi )\} \in U_{\gamma_n}
$$
$$\text{ iff }(\ast)_1 (M, \in, U \rangle \models \{\{\xi_1,\dots,\xi_{n−1}\} \in [κ]^{n−1} \mid \{\xi \lt \cup U \mid \varphi(x, \xi_1,\dots,\xi_{n−1}, \xi )\} \in U\} \in U^{n−1}
$$
$$\text{ iff } (M, \in, U \rangle \models \{\{\xi_1,\dots,\xi_n\} \in [κ]^n \mid \varphi(x, \xi_1,\dots,\xi_n)\} \in U^n.
$$
The step $(\ast)_1$ is now perfectly justified by our inductive hypothesis. In the step $(\ast)_0$, we used Los's theorem for the $\Sigma$-formulas. This finishes the proof.
Best Answer
This is not true without additional hypotheses. For instance, if for some $\lambda<\kappa$ you have a normal $\lambda$-complete ultrafilter $W$ on $\lambda$, then $W$ generates an ultrafilter $U$ on $\kappa$ which will still be normal but is not $\kappa$-complete.
It is true if you additionally assume that every element of $U$ is unbounded. To prove it, let $\lambda<\kappa$ and let $(A_\alpha)_{\alpha<\lambda}\in M$ be a sequence of subsets of $\kappa$ that are not in $U$; we must show that the union $A=\bigcup A_\alpha$ is also not in $U$. Define $f:\kappa\to\kappa$ by letting $f(\xi)$ be the least $\alpha$ such that $\xi\in A_\alpha$ if $\xi\in A$, and $f(\xi)=\lambda$ otherwise. Then $f(\xi)\leq\lambda$ for all $\xi$, and so $\{\xi<\kappa:f(\xi)<\xi\}$ is cobounded and thus in $U$. By normality, there is thus some $\alpha$ such that $\{\xi<\kappa:f(\xi)=\alpha\}\in U$. But if $\alpha<\lambda$ then $\{\xi<\kappa:f(\xi)=\alpha\}\subseteq A_\alpha$ and thus is not in $U$. We must thus have $\alpha=\lambda$ so $\{\xi<\kappa:f(\xi)=\lambda\}=\kappa\setminus A\in U$, as desired.