= Solution
<Krasner's lemma> states: let $K$ be <complete> for a <Non-Archimedean absolute value>, let $\alpha$ be a <separable algebraic element> over $K$, and let $\beta$ be algebraic over $K$, all in a fixed <algebraic closure> with the extended <field absolute value>. If
$$
|\beta-\alpha|<|\alpha'-\alpha|\quad\text{for every }K\text{-conjugate }\alpha'\ne\alpha,
$$
then $K(\alpha)\subseteq K(\beta)$. If $\alpha\in K$, the conclusion is automatic and there is no conjugate-distance condition to check.
To prove it, put $F=K(\beta)$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a <separable polynomial>, so $N/F$ is Galois. The finite extension $F$ is <complete>, and uniqueness of extensions of an <field absolute value> from a <complete> non-Archimedean field makes every $\sigma\in\operatorname{Gal}(N/F)$ an isometry. It fixes $\beta$, so
$$
|\sigma(\alpha)-\alpha|\le\max\{|\sigma(\alpha)-\beta|,|\beta-\alpha|\}=|\beta-\alpha|.
$$
If $\sigma(\alpha)\ne\alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The <Fundamental theorem of Galois theory> gives $\alpha\in F$, proving
$$
\boxed{K(\alpha)\subseteq K(\beta).}
$$
Only $\alpha$ needs to be a <separable algebraic element>: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta)$.
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