This is just a consequence of Nullstellensatz (the flatness is useless). If $x\in X$ is a closed point, then $k(x)$ is a finite extension of $k$ and so is $k(f(x))$ (being a subextension of $k(x)$). This is enough to show that $f(x)$ is a closed point.
If $X, Y$ are not finite type over a field, this is false even if $f$ is finite type (and faithfully flat if you like): consider a DVR $R$ with uniformizing element $\pi$, and let $f : X=\mathbb A^1_R \to Y=\mathrm{Spec}(R)$ be the canonical morphism. It is finite type and faithfully flat. The polynomial $1-\pi T\in R[T]$ defines a closed point of $X$ whose image by $f$ is the generic point of $Y$.
$\newcommand{\spec}[1]{\mathrm{Spec}\,(#1)}$
I'm new at this, so it is more thoughts put into words than an answer.
By definition of a fiber product (in any locally small category for that matter), we have :
$$ X_y(-) = X(-) \times_{Y(-)} \spec{k(y)}(-)$$
(where $Z(-)$ is the functor of points $\hom_{k-\mathbf{Sch}}(-,Z)$ of the $k$-scheme $Z$).
Explicitely, for a $k$-scheme $Z$, $X_y(Z) = \{ (\varphi,\psi) \in X(Z) \times \spec{k(y)}(Z) : f \circ \varphi = i_y \circ \psi \}$ where $i_y$ is the (inclusion) morphism $\spec{k(y)} \to Y$. But then, the functor of points of $\spec{k(y)}$ admits a nice description : any morphism $Z \to \spec {k(y)}$ is topologically the constant map to the single point of $\spec {k(y)}$, and the map of structuring sheaves is just a morphism $k(y) \to \mathcal O_Z(Z)$ of $k$-algebra.
Now take $Z$ to be $\spec A$ for a $k$-algebra $A$, we have :
$$ \spec{k(y)} (A) \cong \hom_{k}(k(y), A). $$
So an element of $X_y(A)$ is the data of a morphism $\spec A \to X$ and a structure of $k(y)$-algebra on $A$ extending the structure of $k$-algebra, such that
- topologically, $\spec A$ factors through $f^{-1}(\{y\})$,
- for every $x \in f^{-1}(\{y\})$, the structure of $k(y)$-algebra of $A$ factors through the residue field $k(x)$ of $x$.
If now $A$ is a $k(y)$-algebra and you're interesting in the $A$-points of $X_y$ as $k(y)$-schemes$^{(1)}$, then the structure of $k(y)$-algebra is imposed but the conditions remain. So I would say that your assumption
Using the universal property of the fiber product, is this equivalent to the $A$-points of $X$ where $A$ is a commutative $k$-algebra via $k \hookrightarrow k(y)$?
was not restrictive enough.
(1) This is not clear in the OP what kind of $A$-points you are looking for.
Best Answer
An integral morphism preserves (and also reflects) closed points. This is because for an integral extension of integral domains $D \subseteq D'$ we know that $D$ is a field iff $D'$ is a field.
Besides, there are many important and well-known strengthenings:
projective $\Longrightarrow$ proper $\Rightarrow$ closed (=preserves closed subsets) $\Rightarrow$ preserves closed points