= Solution
A <morphism of schemes> is <proper morphism>[proper] when it is a <finite type morphism>, a <separated morphism>, and a <universally closed morphism>. The <valuative criterion for properness> says, under the usual finite-type and Noetherian hypotheses, that $f:X\to Y$ is proper exactly when every commutative square
$$
\begin{array}{ccc}
\operatorname{Spec}K&\longrightarrow&X\\
\big\downarrow&&\big\downarrow f\\
\operatorname{Spec}R&\longrightarrow&Y
\end{array}
$$
with $R$ a <discrete valuation ring> and $K=\operatorname{Frac}R$ has a unique diagonal lift $\operatorname{Spec}R\to X$.
For $\mathbb P_k^2\to\operatorname{Spec}k$, a $K$-point is $[x_0:x_1:x_2]$ with not all $x_i$ zero. If $\pi$ is a <uniformizer>, multiply all coordinates by one power of $\pi$ so that $\min_i v(x_i)=0$. The resulting coordinates lie in $R$ and at least one is a <unit in a ring>[unit], so they define an $R$-point extending the given $K$-point. If two extensions exist, on a chart where one coordinate is a unit their affine coordinate ratios agree in $K$ and therefore in the <integral domain> $R$; hence the extensions agree. This verifies existence and uniqueness directly.
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