= Solution
\b[A number-field completion is finite.] Let $F$ be the given <number field> and $L=F_{\mathfrak p}$ its completion. Normalize its <absolute value on a field> so that the restriction to $\mathbb Q$ is $|\cdot|_p$: if $v_{\mathfrak p}(p)=e$, use $|x|=p^{-v_{\mathfrak p}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb Q$ in $L$ is an isometric copy of $\mathbb Q_p$.
Take a $\mathbb Q$-basis of $F$ and let $V$ be its $\mathbb Q_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The <compact-sphere proof of finite-dimensional non-Archimedean norm equivalence> applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence
$$
\boxed{[F_{\mathfrak p}:\mathbb Q_p]\leq[F:\mathbb Q]<\infty.}
$$
\b[Every finite extension is obtained this way.] Let $E/\mathbb Q_p$ be finite. Use its extending <Non-Archimedean absolute value>, whose completeness and uniqueness were proved above. In characteristic zero the <primitive element theorem> gives $E=\mathbb Q_p(\alpha)$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to $\mathbb Z_p[T]$. If the degree is one, $E=\mathbb Q_p$ is already the completion of $\mathbb Q$.
Otherwise choose
$$
0<\delta<\min_{\alpha'\ne\alpha}|\alpha-\alpha'|,
$$
where $\alpha'$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in\mathbb Z[T]$ of the same degree. Since $f'(\alpha)\ne0$, sufficiently accurate approximation gives $|g(\alpha)|<|g'(\alpha)|^2$ and makes $|g(\alpha)/g'(\alpha)|<\delta$. The <strong form of Hensel lemma> in $E$ produces a root $\beta\in E$ of $g$ with $|\beta-\alpha|<\delta$.
For completeness, the required <Krasner's lemma> follows from the <unique extension of an absolute value to a finite extension>. In a finite <Galois extension> containing $\alpha,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, and
$$
|\sigma\alpha-\alpha|
\leq\max(|\sigma\alpha-\beta|,|\beta-\alpha|)
=|\alpha-\beta|.
$$
If $\sigma\alpha\ne\alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and <Galois correspondence> gives $\mathbb Q_p(\alpha)\subseteq\mathbb Q_p(\beta)$. Since $\beta\in E$, equality follows. In particular $g$ is irreducible over $\mathbb Q_p$, and therefore over $\mathbb Q$.
Put $F=\mathbb Q(\beta)\subset E$. It is a <number field>, and the closure of $F$ in $E$ contains $\mathbb Q_p(\beta)=E$, so $F$ is dense. Let
$$
\mathfrak p=\{b\in\mathcal O_F:|b|<1\}.
$$
Every element of the <ring of integers of a number field> has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the <residue field> of $E$ makes $\mathfrak p$ a <prime ideal>, and $\mathfrak p\cap\mathbb Z=p\mathbb Z$. The restriction of the <valuation> to $F$ is the <valuation> of this prime, up to normalization. Indeed, the localized <discrete valuation ring> $(\mathcal O_F)_{\mathfrak p}$ maps into $\mathcal O_E$, its units have absolute value one, and its <uniformizer> has absolute value less than one. Density and completeness give
$$
\boxed{F_{\mathfrak p}\cong E.}
$$
This proves the <local realization as a completion of a number field>.
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