= Solution
Write $q=\#k_K=p^f$. For every $a\in k_K$, the polynomial $F(T)=T^q-T$ has $a$ as a residue root and $\overline{F'}=-1$. By the <Hensel lemma> there is a unique element $[a]\in\mathcal O_K$ such that
$$
\boxed{\overline{[a]}=a,\qquad [a]^q=[a].}
$$
This defines the <Teichmuller lift>, including $[0]=0$. The two displayed properties determine each value uniquely by Hensel's uniqueness statement.
They also imply the standard <Teichmuller lift properties>. The elements $0,1$ already satisfy their defining equations, so $[0]=0$ and $[1]=1$. The product $[a][b]$ reduces to $ab$ and satisfies $( [a][b])^q=[a][b]$, so uniqueness gives $[ab]=[a][b]$. Likewise $[a]^p$ reduces to $a^p$ and is fixed by its $q$th power, giving
$$
\boxed{[ab]=[a][b],\qquad [a^p]=[a]^p.}
$$
For $a\ne0$, division of $[a]^q=[a]$ by $[a]$ gives $[a]^{q-1}=1$. Thus the nonzero <Teichmuller representatives> are precisely the prime-to-$p$ roots of unity lifting the nonzero residue classes. This is a multiplicative section of reduction; additivity is not asserted in mixed characteristic.
For the convergence, use the normalized <discrete valuation> $v_K$ with $v_K(\pi)=1$, and put $e=v_K(p)\geq1$. A useful binomial estimate is
$$
v_K(u^p-v^p)\geq\min\{e+r,pr\}\geq r+1
\quad\text{if }u,v\in\mathcal O_K,\ v_K(u-v)\geq r\geq1.
$$
Indeed, writing $u=v+\delta$, every intermediate coefficient $\binom pj$ is divisible by $p$, so all terms of $(v+\delta)^p-v^p$ have valuation at least $\min\{e+r,pr\}$. The <ultrametric inequality> gives the bound for their sum.
Finite fields are perfect, so the successive $p$th roots $x_n$ exist uniquely and satisfy $x_n^{p^n}=x$. Let $a_n=[x_n]$. Since $y_n-a_n$ lies in the maximal ideal, iterating the estimate $n$ times gives
$$
v_K(y_n^{p^n}-a_n^{p^n})\geq n+1.
$$
Multiplicativity, or the Frobenius identity above, gives $a_n^{p^n}=[x_n^{p^n}]=[x]$. Therefore
$$
\boxed{v_K(y_n^{p^n}-[x])\geq n+1,\qquad y_n^{p^n}\longrightarrow[x].}
$$
This proves the <approximation of Teichmuller lifts by iterated pth powers> for completely arbitrary choices of the lifts $y_n$, including $x=0$. The original PDF confirms $x\in k_K$, correcting the capital $K_K$ in the TeX transcription.
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