I might miss the forest for the trees but:
The situation is (too) special, I think. Nevertheless I'll describe it: I have an open morphism $X\to Y$ of schemes (morphism is locally of finite type). The schemes are locally Noetherian and $Y$ is irreducible with generic point $\eta$. Why (due to the openness of $f$) do we have $f(\xi)=\eta$ for any generic point of $X$?
The more general question (as I do think we don't need the assertions above): Why does an open continuous map into an irreducible space send generic points of the domain to the generic point of the target space?
Thank you!!
Assuming that by "generic point of $X$" you mean "generic point of an irreducible component of $X$", this is true for any locally Noetherian space $X$. First, note that we may assume $X$ is irreducible: let $U$ be a Noetherian open set containing $\xi$, and then restrict $f$ to the complement in $U$ of all of the irreducible components of $U$ except the closure of $\xi$ (this complement is still open since there are only finitely many such components, by Noetherianness).
Now if $X$ is irreducible, every point in $X$ is in the closure of $\{\xi\}$. If $f:X\to Y$ is continuous, that means every point in $f(X)$ is in the closure of $\{f(\xi)\}$. If $f$ is also an open map, this means that the closure of $\{f(\xi)\}$ has nonempty interior. If $Y$ is irreducible, this means the closure of $\{f(\xi)\}$ must be all of $Y$, since every nonempty open set is dense in $Y$. Thus $f(\xi)$ must be the generic point of $Y$.