= Solution
At a point $P$ of the <smooth algebraic curve> $X$, the <local ring> $\mathcal O_{X,P}$ is a <discrete valuation ring>. Embed the <projective variety> $Y$ in $\mathbb P^N$. A <rational map of algebraic varieties> to $Y$ has homogeneous coordinates $[f_0:\cdots:f_N]$ in the <function field> $k(X)$, not all zero. Let $v_P$ be the local <valuation>, choose a <uniformizer> $u$, and set $m=\min_{f_i\ne0}v_P(f_i)$. The coordinates $u^{-m}f_i$ are regular at $P$, and at least one is a unit. On a sufficiently small <open subset> containing $P$, all these functions are regular and that coordinate stays nonzero. Their ratios therefore define a <morphism of algebraic varieties> into the corresponding affine chart of $\mathbb P^N$. Every homogeneous equation defining $Y$ vanishes on the original dense open, hence vanishes identically on this neighborhood. The local extension consequently lands in $Y$.
Two such extensions agree wherever the original <rational map of algebraic varieties> is defined. Their equalizer is closed, because $Y$ is a <separated variety>, and contains a dense open. Thus they agree throughout their overlap and glue. \b[The rational map has a unique extension to a morphism on all of $X$.] Projectivity is needed for the target; the source need not be projective.
For the <function field> correspondence, let $K/k$ be a finitely generated <field extension> of <transcendence degree> one. Choose a <transcendence basis> consisting of one element $t$. The extension $K/k(t)$ is finite: finitely many algebraic generators each have finite degree. Take the <integral closures> of $k[t]$ and $k[t^{-1}]$ in $K$. The finiteness assumption allowed in the paper makes both finite modules over their original coordinate rings. <Integral closure> commutes with localization, so these two rings agree over $k[t,t^{-1}]$ and glue to the <finite normalization model of a one-variable function field> $C_K\to\mathbb P^1$. This is a <finite morphism>, hence projective, and its <function field> is $K$. Its one-dimensional normal local rings are <discrete valuation rings>, so it is regular. Since the <algebraically closed field> $k$ is perfect, regularity here implies smoothness. Thus $C_K$ is a <smooth projective curve>.
Conversely the <function field> of a <smooth projective curve> is finitely generated over $k$ and has <transcendence degree> one. Two such curves with isomorphic <function fields> have mutually inverse <birational maps>. The extension argument just proved makes both maps morphisms; their composites equal the identity on a dense open, and hence everywhere by <separatedness>. This proves uniqueness of the smooth projective model. More precisely, pullback gives a contravariant correspondence between nonconstant morphisms of <smooth projective curves> and $k$-embeddings of their <function fields>. A field embedding defines the map on affine coordinates near the <generic point>, and the extension argument completes it. Constant morphisms are excluded from this correspondence because they do not induce embeddings of <function fields>.
Finally let $f:C\to D$ be nonconstant. Its image is dense, and pullback gives a finite extension $k(D)\subset k(C)$: both fields have <transcendence degree> one and the latter is finitely generated. Normalize $D$ in $k(C)$, applying the allowed finite-<integral closure> result on affine charts. This gives a <finite morphism> $C'\to D$ with $C'$ a <smooth projective curve> and <function field> $k(C)$. The identification of <function fields> gives an isomorphism $C\cong C'$ by the preceding paragraph. Its composite with $C'\to D$ equals $f$ at the <generic point>, hence equals $f$ everywhere. \b[Every nonconstant morphism of these curves is finite.]
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