= Solution
<Krasner's lemma> states that if $K$ is complete, $\alpha$ is separable over $K$, and an algebraic element $\beta$ satisfies
$$
|\beta-\alpha|<|\alpha'-\alpha|
$$
for every other $K$-conjugate $\alpha'$ of $\alpha$, then $K(\alpha)\subseteq K(\beta)$.
Let $f\in\mathbb C_p[X]$ be nonconstant and let $\alpha$ be a root in an algebraic closure of $\mathbb C_p$. Because the characteristic is zero, replace $f$ by the separable minimal polynomial of $\alpha$. Approximate its coefficients arbitrarily closely by elements of $\overline{\mathbb Q}_p$. By <continuity of roots over a non-Archimedean field>, the approximating polynomial has a root $\beta$ arbitrarily close to $\alpha$. Since $\overline{\mathbb Q}_p$ is algebraically closed, $\beta\in\overline{\mathbb Q}_p\subset\mathbb C_p$.
Choose the approximation so that $\beta$ is closer to $\alpha$ than every other $\mathbb C_p$-conjugate of $\alpha$. Krasner's lemma gives
$$
\mathbb C_p(\alpha)\subseteq\mathbb C_p(\beta)=\mathbb C_p,
$$
so $\alpha\in\mathbb C_p$. Hence <completion of an algebraic closure of a p-adic field is algebraically closed> proves that \b[$\mathbb C_p$ is algebraically closed.]
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