= Solution
Suppose first that the valuation is discrete and the residue field $k$ is finite. For a uniformizer $\pi$, every quotient $\mathcal O_K/\pi^n\mathcal O_K$ is finite, and completeness gives
$$
\mathcal O_K\cong\varprojlim_n\mathcal O_K/\pi^n\mathcal O_K.
$$
This inverse limit is compact. Since $\mathcal O_K$ is a compact neighborhood of zero, $K$ is locally compact.
Conversely, local compactness gives a compact ball about zero, which can be rescaled to make $\mathcal O_K$ compact. Its distinct residue classes are disjoint open balls of radius below one, so compactness forces the residue field to be finite. Cover $\mathcal O_K$ by finitely many balls of some radius $r<1$. Applying the <ultrametric inequality> to centers lying in the maximal ideal produces $\rho<1$ such that every nonunit has absolute value at most $\rho$. Hence the value group has a largest value below one, and the valuation is discrete. This proves the <local compactness criterion for a complete non-Archimedean field>.
An algebraically closed valued field has an $n$th root of every element. If its valuation were discrete and $\pi$ were a uniformizer, then $v(\sqrt[n]{\pi})=1/n$ would contradict discreteness. Therefore an algebraically closed non-Archimedean field cannot be locally compact.
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