If $N$ is a submodule of $M$, then for any $x$ such that $xM = 0,$ then given arbitrary $n ∈ N, n ∈ M, xn = 0.$ Then the collection of $x$'s that annihilate $M$ contain the collection of $x$'s that annihilate $N$ because $xm=0$ implies $xn=0.$ So $\mathrm{Ann}(N) \subseteq \mathrm{Ann}(M).$
It reminds me of the following:
If $A$ and $B$ sets, then $A \subseteq B$ implies $\mu^{*}(A) \leq \mu^{*}(B).$ Because:
If both $A$ and $B$ have a cover, since $A \subseteq B,$ any cover of $B$ is also a cover of $A$. So the collection of covers of $B$ is contained in the collection of covers of $A$. That means that there are more collections to consider when computing $m^*(A)$ instead of $m^*(B)$. So by the definition of infimum, we have $m^*(A)\le m^*(B).$
I feel like this the top paragraph is right, but am not positive.
What do you think?