Proof outer measure satisfies monotonicity: $A \subseteq B \implies m^*(A) \leq m^*(B)$

I am just a little bit unsighted by your proof, hence I want to give you a proof which is roughly the same, but a little more sharpened so that you can see how get rid of small discrepancies in proofs.

So $m^*(B) = \inf \{ \sum_1^n|J_k|, \{ J_k\} \text{ is an open interval cover of B}\}$.

Similarly, $m^*(A) = \inf \{ \sum_1^n|J_k|, \{ J_k\} \text{ is an open interval cover of A}\}$.

Note that $A \subset B$,whence if $\{J_k\}$ covers $B$, it also covers $A$. Hence, the following is true: $$ \{ \sum_1^n|J_k|, \{ J_k\} \text{ an open interval cover of B}\} \subset \{ \sum_1^n|J_k|, \{ J_k\} \text{ an open interval cover of A}\} $$

Now, the infimum of the subset of a family is larger than the infimum of the family, because the family could contain an element that is not in the subset but smaller than every element in the subset.

Hence, it follows that the infimum of the left side, which is $m^*(B)$, is greater than the right side, which is $m^*(A)$.

EDIT: As other have pointed out, there was something wrong in your proof. However, you were thinking in the right direction, and hopefully this proof will show you how to put those efforts in a more rigorous direction.