= Solution
Let $\mathcal U$ be a <nonprincipal ultrafilter> on an infinite set $I$. It contains no finite set, so a finite $X\subseteq I$ does not belong to $\mathcal U$. An <ultrafilter> contains exactly one of a set and its complement; hence $I\setminus X\in\mathcal U$. Equivalently, every nonprincipal ultrafilter contains the <cofinite filter>.
For structures $(M_i)_{i\in I}$ and a first-order formula $\varphi$, the <Łoś theorem> says
$$
\prod_{i\in I}M_i/\mathcal U\models\varphi([a_i^1],\ldots,[a_i^k])
\quad\Longleftrightarrow\quad
\{i:M_i\models\varphi(a_i^1,\ldots,a_i^k)\}\in\mathcal U.
$$
Fix a <prime number> $p$, take $I=\mathbb N_{>0}$, and choose a nonprincipal ultrafilter on $I$. The <ultraproduct>
$$
K=\prod_{n\geq1}\mathbb F_{p^n}/\mathcal U
$$
is a <field> because each factor is a <finite field>, and it has <characteristic of a field>[characteristic] $p$ because each factor satisfies $p\cdot1=0$ and $m\cdot1\ne0$ for $1\leq m<p$. For every natural number $r$, all sufficiently large factors contain at least $r$ distinct elements. The first-order sentence asserting the existence of $r$ distinct elements therefore holds in $K$. Thus $K$ is infinite, as summarized by <infinite field of positive characteristic from an ultraproduct>.
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