Let $A$ and $B$ be two rings. We have a ring morphism $f: A\to B$. We denote $\phi: \newcommand{\Spec}{\operatorname{Spec}}\Spec B \to \Spec A$ the morphism induced by $f$ between Spec. Then we have:
$\phi$ is open $\iff$ if $\exists J \in \Spec A : J \subseteq \phi(\mathfrak{p})$ then there exists $\mathfrak{p_{j}} \in \Spec B : \mathfrak{p_{j}} \subseteq \mathfrak{p}$ and $\phi(\mathfrak{p_{j}})=J$.
Is that correct? How I can prove it?
The condition you describe (I guess the name for it is "going down property") will imply that $\phi(\mathrm{Spec}\,B)$ is "generization closed". It will not be an open set in general. In particular, $\phi$ will not be open in general.
As an example, consider $f$ to be the localization $A \rightarrow A_{\mathfrak{p}}$ for a prime ideal $\mathfrak{p}$. This will always satisfy your property (by correspondende of primes in a localization) but usually it will not be open. As a concrete instace, take $\mathbb{Z} \rightarrow \mathbb{Z}_{(2)}$. Then $\phi(\mathbb{Z}_{(2)})=\{[(0)], [(2)]\}$ so this is not an open set.
EDIT: However, it is true under the assumption that $A \rightarrow B$ is of finite presentation, and it is also true in general that open maps satisfy the going down property. See Stacks Project, tags 0407 and 00I1 .