= Solution
The intended statement uses a nontrivial <Non-Archimedean absolute value>. Under that convention choose $t$ with $0<|t|<1$. A compact neighborhood $C$ of zero contains an open ball of some radius $r>0$. For sufficiently large $N$, $t^N\mathcal O_K$ is contained in that ball. The <valuation ring> is closed, so $t^N\mathcal O_K$ is a closed subset of $C$ and is compact. Scaling shows that $\mathcal O_K$ itself is compact.
Its <maximal ideal> $\mathfrak m=\{x:|x|<1\}$ is open. The <residue field> $k=\mathcal O_K/\mathfrak m$ is therefore discrete and compact, hence finite: the open cover by its singleton sets has a finite subcover. Since $\mathfrak m$ has finite index, its complement is a finite union of open cosets. Thus $\mathfrak m$ is also closed and compact.
The continuous <absolute value on a field> attains its maximum on $\mathfrak m$. Nontriviality supplies a nonzero element there, so this maximum is $\rho=|\pi|$ for some $\pi\ne0$, with $0<\rho<1$. If $x\in\mathfrak m$, then $|x/\pi|\leq1$, so $\mathfrak m=\pi\mathcal O_K$.
For arbitrary $x\ne0$, choose the integer $n$ for which $\rho^{n+1}<|x|\leq\rho^n$. Then $\rho<|x/\pi^n|\leq1$. This element cannot lie in $\mathfrak m$, by maximality of $\rho$, so its <absolute value on a field> is one. Therefore $|x|=\rho^n$. This proves <discrete valuation from nontrivial local compactness>:
$$
\boxed{|K^\times|=\rho^{\mathbb Z},\qquad k\text{ finite}.}
$$
The <nontrivial valuation in the local compactness criterion> is essential. If the <trivial absolute value> is admitted, the displayed source claim is false: take $\mathbb Q$ with that <absolute value on a field>. It is complete and discrete, hence locally compact, but its <valuation ring> and <residue field> are both $\mathbb Q$, which is infinite. The intended nontrivial case has been proved above.
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