I'm struggling a lot on this problem.
This feels intuitionistically true, since if their probabilities are equal, that means that the joint probability did not affect the size of the sample space that p maps to, which must mean that A is a subset.
How would I derive this from axioms?
You have $P(A)-P(A\cap B)=P(A\cap B^c)=0$, so $\omega\notin A\cap B^c$ for $P$-almost all $\omega$, which means that $A^c\cup B$ contains $P$-almost all $\omega$, or in other words, $(\omega\in A\implies\omega\in B)$ holds $P$-almost surely. By a slight abuse of language, one says that $A$ is almost surely included in $B$.