So lets take two open sets $U,V \subseteq Y \subseteq X$, where $Y$ is open as well.
That means we have two morphisms $U \to Y$, $V \to Y$.
Now the fibered product of those morphisms (lets call it $W$) is an object of the same category - hence it is an open subset of $X$.
Furthermore we need some maps $W \to U$, $W \to V$.
By the definition of our category this translates into: $W \subseteq U$, $W\subseteq V$. Hence we have $W \subseteq U\cap V$ for sure.
Now we want the following: Given any open subset $W' \subseteq X$ together with morphisms $W' \to U$, $W'\to V$ that make the appropriate diagram commute (which is trivial in our case, do you see why?), we want a morphism $W' \to W$ that makes all "new" diagrams commute (again, it is trivial that all diagrams commute once we have a morphism - that is all we need to check).
$W' \to W$ translates to $W' \subseteq W$. Hence intuitively we'd expect $W$ to be the biggest open set contained in both $U, V$. As you've already said, that is $U \cap V$. Now why does $U \cap V$ satisfy the universal property of the fiber product? Because if $W' \subseteq U$, $W' \subseteq V$ it follows that $W' \subseteq U \cap V$, which exactly means that there is a morphism $W' \to U\cap V$. Again, by the very definition of our category it is clear, that this morphism is unique - there is at most one morphism between any two objects.