Proof of $m^*(\mathbb{R}^n)=\infty.$ Let $\mathbb{R^n}$ whole set.
For $\displaystyle E\subset \mathbb{R^n},$ define Lebesgue outer measure $m^*$ as $$m^*(E)=\inf \bigg\{ \sum_{k=1}^{\infty} |I_k| \ \bigg| \ E\subset \bigcup_{k=1}^{\infty} I_k \bigg\}$$ where  each $I_k$ is a $n$-dimentional rectangle and $|I_k|$ stands for the volume of $I_k$.
Then, prove that $m^*(\mathbb{R}^n)=\infty.$
My proof
Let $M>0$ arbitrary.
Then, set $L:=[-\dfrac{M^{\frac{1}{n}}}{2},\dfrac{M^{\frac{1}{n}}}{2}]\times [-\dfrac{M^{\frac{1}{n}}}{2},\dfrac{M^{\frac{1}{n}}}{2}]\times \cdots \times [-\dfrac{M^{\frac{1}{n}}}{2},\dfrac{M^{\frac{1}{n}}}{2}]$ and I get $m^*(L)=M.$
Since $L \subset \mathbb{R^n}$, $M=m^*(L)\leqq m^*(\mathbb{R}^n)$ from the monotonicity of $m^*.$
Since $M$ is arbitrary positive, $m^*(\mathbb{R}^n)=\infty.$
Is this proof correct ?

Assuming you've shown $m^*$ is monotonic, your proof looks fine.
Note that there are simpler candidates for $L=L_M$ unless you're wanting $\bigcup_{M>0} L_M = \mathbb{R}^n$ (which yours does).  Specifically, you could just use $$L = [0,M]\times [0,1]^{n-1}$$ The benefit being that it's easier to write and the calculation of $m^*(L) = M$ is almost trivial.
