Let $f: C \to D$ a finite morphism between two $1$-dimensional integral and proper schemes $C$ and $D$ over field $k$ (therefore $C \to Spec(k)$ and $D \to Spec(k)$ proper). Let $\eta_C$ and $\eta_D$ generic points of $C$ and $D$.
I want to see why $f(\eta_C) =\eta_D$ holds.
My attempts: Let $a:= f(\eta_C)$. As $D$ $1$-dimensional and irreducible (since integral) each point of $D$ is closed or generic $\eta_D$. Let assump $a$ not generic, so closed. Therefore $A := f^{-1}(a)$ is closed and contains $\eta_C$, so $A = C$.
Therefore $f(C)= \{a\}$ and $f$ factors throught a point. But I don't see how to deduce a contradiction to to hypothesis.
If $f$ is finite, then take any affine open $V = Spec(B)$ in $D$ for which $f^{-1}(V) = Spec(A) \subset C$, then look at the restriction $f: Spec(A) \to Spec(B)$ which corresponds to a $k$-algebra homomorphism $B \to A$ making $A$ a finitely generated $B$-module (we can do this by definition of $f$ being affine). Note $A$ and $B$ are $1$-dimensional integral domains by assumption, and the generic points of $C$ and $D$ are the zero ideals in $A$ and $B$. You want to show that the preimage of the zero ideal in $A$ is the zero ideal in $B$, i.e. you want $B \to A$ to be injective.
If the preimage of the zero ideal in $A$ is some nonzero prime $\mathfrak p$ then $B/\mathfrak p \hookrightarrow A$ will still make $A$ a finitely generated $B/\mathfrak p$ module, but now $B/\mathfrak p$ is a field (some finite extension $K$ of $k$) by the dimension assumption. But this contradicts Noether normalization, which tells you that $A$ has transcendence degree $1$ over $K$ (or $k$ really, but $K$ is finite over $k$).