Let $A,B$ $\subseteq \mathbb{R}$ such that $A\subseteq B$ then
$\mu^*(A)\leq \mu^*(B)$. Where $\mu^*$ is the Lebesgue measure.
My Attempt:
Let $I_1,I_2,I_3...$ be a sequence of open intervals which cover $B$. Then they cover $A$ thus
$\bigl\{\sum^{\infty}_{k=1}l(I_k): I_1,I_2...$ are open intervals with $B\subseteq \bigcup_{k=1}I_k \bigr\} \subseteq\{$ $\sum^{\infty}_{m=1}l(I_m): I_1,I_2...$ are open intervals with $\;A\subseteq \bigcup_{m=1}I_m \bigr\}$.
Using the result that $A\subseteq B$ $\implies$ $\inf B\leq \inf A$, I conclude that $\mu^*(A)\leq \mu^*(B)$
Is my attempt correct?
Yes, this is correct. We know that for $A\subseteq \mathbb{R}$, $\mu^*(A)=\inf\{\sum_k \ell(I_k): \{I_k\}\:\text{is a countable cover by open intervals of A}\}.$ If $A\subseteq B$, then cover of $B$ by open intervals is an open cover of $A$ by open intervals. Moreover, if $$\Omega_A=\{ \sum_k \ell(I_k):\{I_k\}\:\text{is a countable cover by open intervals of A}\}$$ $$ \Omega_B=\{ \sum_k \ell(I_k):\{I_k\}\:\text{is a countable cover by open intervals of B}\}$$ then as you say, $\Omega_A\supseteq \Omega_B$. So, $\inf \Omega_A\le \inf \Omega_B$. Hence $\mu^*(A)\le \mu^*(B)$.