 Solution

One form of Krasner's lemma is: if $K$ is a complete non-Archimedean field, $\alpha$ is separable over $K$, and $\beta$ is algebraic withwhere ${\alpha }^{\prime }$ ranges over the other $K$-conjugates, then $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is immediate.

To prove the nontrivial case, let $\sigma$ be an automorphism of an algebraic closure fixing $K(\beta )$. The unique extension of an absolute value to a finite extension implies that $\sigma$ preserves the extended absolute value. ThereforeThis is strictly less than the distance to every other conjugate, so $\sigma (\alpha )=\alpha$. As $\alpha$ is separable over $K(\beta )$, being fixed by all these automorphisms puts it in $K(\beta )$. This proves $\boxed{K(\alpha )\subseteq K(\beta )}$. The strict inequality is what prevents switching to another conjugate.