I want to prove the claim:
$\bf{Proposition}$ If $E\subset F \subset \mathbb{R}^d$, then $m^*(E)\leq m^*(F)$, where $m^*(S)$ denotes Lebesgue outer measure of set $S$.
I tried to prove it by using "proof of contradiction" as below. Could you check this?
$\bf{Proof}$Assume, for the sake of contradiction, that $E\subset F\Rightarrow m^*(E)>m^*(F)$. Due to the monotonicity of the Jordan outer measure, we can say $m^{*,(J)}(E)<m^{*,(J)}(F)$. Due to the relation of Jordan and Lebesgue outer measure, we can say $m^{*}(E)\leq m^{*,(J)}(E)$. Hence, using the assumption, $m^*(F)<m^*(E)\leq m^{*,(J)} \leq m^{*,(J)}(F)$. However, this contradicts the relation of Jordan and Lebesgue outer measure for $F$: $m^*(F)\leq m^{*,(J)}(F)$, so the claim follows. $\square$
Here are some critiques:
1) You didn't negate the statement correctly. You should assume for the sake of contradiction that there exist sets $E,F$ with $E \subset F$ and $m^*(E) > m^*(F)$. There is no implication.
2) I think the proper term for $m^{*,(J)}(E)$ is the Jordan outer content.
3) The best you can do with the Jordan outer content is $m^{*,(J)}(E) \le m^{*,(J)}(F)$. You don't have strict inequality.
4) You concluded that $m^*(F) < m^{*,(J)}(F)$. This is in no way whatsoever contradictory with $m^*(F) \le m^{*,(J)}(F)$.
For example, $1 < 2$ does not contradict $1 \le 2$.
Points 1-3 are minor and easily corrected. Point 4 is serious and can't be easily corrected.