$(\mathfrak{p}+(x)).(\mathfrak{p}+(y))\subseteq\mathfrak{p}+(xy)$ is rather straightforward to show considering the elements of the ideals on the left and definition of sum and product of ideals. Here $\mathfrak{p}$ is a prime ideal of a commutative ring $A$, and $A\ni x,y\not\in \mathfrak{p}$.
I was wondering if I could lift this problem to the quotient ring $A/\mathfrak{p}$ and then use that $(a).(b)=(ab)$ and use the ring/ideal correspondence theorem to get back to this problem in ring $A$ and conclude something like $(\mathfrak{p}+(x))(\mathfrak{p}+(y))=\mathfrak{p}+(xy)$. Here is what I have tried.
$\pi(\mathfrak{p}+(x))=(\mathfrak{p}+(x))/\mathfrak{p}$ where $\pi:A\to A/\mathfrak{p}$ is the canonical map. An element in $(\mathfrak{p}+(x))/\mathfrak{p}$ is of the form $ax+\mathfrak{p}=a(x+\mathfrak{p})$ for some $a\in A$. Hence, $(\mathfrak{p}+(x))/\mathfrak{p}=(x+\mathfrak{p})\subseteq A/\mathfrak{p}$. Similarly, $\mathfrak{p}+(y)$ gets mapped to $(y+\mathfrak{p})\subseteq A/\mathfrak{p}$. So, $(x+\mathfrak{p}).(y+\mathfrak{p})=(xy+\mathfrak{p})$.
Since $\pi^{-1}(\pi(I))=I+\mathfrak{p}$ for any $I\subseteq A$ ideal, I want to take (not sure if this makes sense) $\pi^{-1}$ throughout $(x+\mathfrak{p}).(y+\mathfrak{p})=(xy+\mathfrak{p})$ and claim that $(\mathfrak{p}+(x)).(\mathfrak{p}+(y))=\mathfrak{p}+(xy)$.
I think my confusion is about how to go from $A$ to $A/\mathfrak{p}$ and then come back to $A$ again. I am still very wobbly in using the correspondence theorem and playing around with $\pi$, I would really appreciate if someone can point out any mistakes.