= Solution
Write the finite <residue field> as $k=\mathbb F_q$, with $q=p^f$ and residue characteristic $p$. Every $\bar a\in k$ satisfies $\bar a^q=\bar a$. The polynomial $F(X)=X^q-X$ has derivative $qX^{q-1}-1$, whose reduction is $-1$. Hence <Hensel lemma> gives a unique lift $[\bar a]\in\mathcal O_K$ such that
$$
[\bar a]\bmod\mathfrak m_K=\bar a,\qquad[\bar a]^q=[\bar a].
$$
This includes $[0]=0$ and $[1]=1$. If $\bar a\ne0$, its lift is a unit and satisfies $[\bar a]^{q-1}=1$.
The product $[\bar a][\bar b]$ also satisfies $X^q=X$ and has residue $\bar a\bar b$. Uniqueness therefore gives $[\bar a\bar b]=[\bar a][\bar b]$. Reduction is a left inverse of this map, so the map is injective. Restricting to nonzero residues proves the requested multiplicative <Teichmuller lift>:
$$
\boxed{k^\times\hookrightarrow\mathcal O_K^\times\subset K^\times,\qquad\bar a\longmapsto[\bar a].}
$$
Its image is the group of $(q-1)$st roots of unity in $K$. Indeed a root of unity of that order is a unit satisfying $X^q=X$, hence is the unique lift of its residue. The only finite-field fact used here is $a^q=a$; for $a\ne0$ it follows from Lagrange's theorem in the multiplicative group of order $q-1$.
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