I think Fact 1 is false.

Let $X=Y$ be any irreducible sober space with generic point $\eta$. Let $f : X\to Y$ be a surjective continuous map and consider the graph $\Gamma_f \subset X\times Y$ of $f$. It is homeomorphic to $X$, hence irreducible. If $\Gamma_f$ has a generic point $(x_0, x_1)$, as $\Gamma_f$ projects onto the two factors of $X\times X$, then $x_0=x_1=\eta$. So $\Gamma_f$ is the closure of $\{ (\eta, \eta)\}$. This implies that $\Gamma_f$, hence $f$, are independent of $f$. So there is only one surjective continuous map $X\to X$, which is obviously fase in general. This also shows that a product $X\times Y$ is always never sober.

Note that in algebraic geometry, the (fiber) product is sober. There is no contradiction with the above reasonning because $(\eta, \eta)$ doesn't correspond to one point in the product, but a big subspace.

Proof of Fact 1 (for any product).

Let $\{ X_i \}_{i\in I}$ be a family of non-empty sober spaces. Let $F$ be a closed irreducible subset of $X:=\prod_i X_i$. By replacing $X_i$ with the closure of the projection of $F$ in $X_i$, we can suppose $F\to X_i$ has dense image for all $i$.

I claim that $F=X$. If $\eta_i$ is the generic point of $X_i$, then it is clear that $(\eta_i)_i$ is the generic point of $X$. So let's prove the claim. Suppose the open subset $X\setminus F$ is non-empty. Then it contains a product $$X\setminus F \supseteq U_{i_1}\times \cdots \times U_{i_n} \times \prod_{i\ne i_1,..., i_n} X_i$$ with non-empty open subsets $U_{i_j}\subseteq X_{i_j}$. So $F$ is covered by finitely many closed subsets of $X$: $$ F\subseteq \cup_{1\le j\le n} Z_{i_j}\times \prod_{i\ne i_j} X_i$$ where $Z_{i_j}=X_{i_j}\setminus U_{i_j}$. As $F$ is irreducible, it is contained in one of them, say $$ F\subseteq Z_{i_1}\times \prod_{i\ne i_1} X_i.$$ But then the projection of $F$ to $X_{i_1}$ is not dense. Contradiction.

I think Fact 1 is false.

Let $X=Y$ be any infinite irreducible sober space with generic point $\eta$. Let $f : X\to Y$ be a surjective continuous map and consider the graph $\Gamma_f \subset X\times Y$ of $f$. It is homeomorphic to $X$, hence irreducible. If $\Gamma_f$ has a generic point $(x_0, x_1)$, as $\Gamma_f$ projects onto the two factors of $X\times X$, then $x_0=x_1=\eta$. So $\Gamma_f$ is the closure of $\{ (\eta, \eta)\}$. This implies that $\Gamma_f$, hence $f$, are independent of $f$, which is obviously fase ($X$ is infinite).

Note that in algebraic geometry, the (fiber) product is sober. Theres no contradiction with the above reasonning because $(\eta, \eta)$ doesn't correspond to one point in the product, but a big subspace.

I think Fact 1 is false.

Let $X=Y$ be any irreducible sober space with generic point $\eta$. Let $f : X\to Y$ be a surjective continuous map and consider the graph $\Gamma_f \subset X\times Y$ of $f$. It is homeomorphic to $X$, hence irreducible. If $\Gamma_f$ has a generic point $(x_0, x_1)$, as $\Gamma_f$ projects onto the two factors of $X\times X$, then $x_0=x_1=\eta$. So $\Gamma_f$ is the closure of $\{ (\eta, \eta)\}$. This implies that $\Gamma_f$, hence $f$, are independent of $f$. So there is only one surjective continuous map $X\to X$, which is obviously fase in general. This also shows that a product $X\times Y$ is always never sober.

Note that in algebraic geometry, the (fiber) product is sober. There is no contradiction with the above reasonning because $(\eta, \eta)$ doesn't correspond to one point in the product, but a big subspace.