Suppose that $\kappa$ is a measurable cardinal, and $U$ is a normal ultrafilter on $\kappa$. Let $X\in U$ be a set. I'm trying to prove that--
$$ \{ \alpha <\kappa \colon \ \alpha \mbox{ is a regular cardinal and } X\cap \alpha \mbox{ is stationary in } \alpha \} \in U$$
I know this can be solved by applying Los' Theorem and considering the ultrapower (see solution below). Is there another way to prove this, without using the ultrapower construction? (we didn't learn ultrapowers and Los' theorem in the course where this problem was given).
(Solution: If $j$ is the ultrapower embedding, it suffices to prove that $j(X)\cap \kappa$ is stationary in $\kappa$, and this holds since $j(X)\cap \kappa =X$ and $X$ is stationary as it belongs to a normal ultrafilter).
Edit: I think that the following argument solves the problem: Denote- $$A = \{ \alpha <\kappa \colon \ \alpha \mbox{ is a regular cardinal and } X\cap \alpha \mbox{ is not stationary in } \alpha \} $$ and assume for contradiction that $A\in U$. For every $\alpha \in A$, let $C_\alpha \subseteq \alpha$ be a club disjoint from $X$.
Define- $$C = \{ \alpha<\kappa \colon \{ \xi \in A \colon \alpha\in C_\xi \}\in U \ \}$$
Clearly, $C,X$ are disjoint. Let us prove that $C$ is a club in $\kappa$, and this is a contradiction.
It suffices to prove that $\{ \alpha \in A \colon C\cap \alpha = C_\alpha \} \in U \ $ (From this it's not hard to verify that $C$ is a club in $\kappa$. We use the fact that each $C_{\alpha}$ is a club in $\alpha$, and therefore $C$ is closed as "high" as we desire).
Claim: $\{ \alpha \in A \colon C\cap \alpha = C_\alpha \} \in U$.
Proof: Otherwise, $\{ \alpha \in A \colon C\cap \alpha \neq C_\alpha \} \in U$. For every $\alpha$ in this set, let $x_{\alpha}$ be an ordinal below $\alpha$ which belongs to exactly one of the sets $C_{\alpha}, C\cap \alpha$. Then $\alpha \mapsto x_{\alpha}$ is regressive, and therefore constant on a set in $U$. Assume that $x$ is the constant value. Then: $$x\in C \iff \{\xi <\kappa \colon x\in C_\xi \}\in U \iff \{ \xi <\kappa \colon x\notin C\cap \xi \} \in U \iff x\notin C$$