 Solution

The disjoint decomposition into sets $A=(A\cap B)\cup (A\setminus B)$ uses only events in $\mathcal{F}$. Finite additivity and nonnegativity giveThus $\mathbb{P}(A\cap B)\le \mathbb{P}(A)$ already follows from the two stated requirements. Equality is attained, for example, by choosing $B=A$, since then $A\cap B=A$. More generally it holds whenever $\mathbb{P}(A\setminus B)=0$. Neither normalization nor countable additivity is needed for this particular inequality.