Let $E\subset \mathbb{R}$ and let $x_0\in \mathbb{R}$
Prove: $m^*(E+x_0)=m^*(E)$
Let $I_n$ be open intervals such that $E\subset \cup_{n=1}^{\infty}I_n$ therfore $E+x_0$ is covered by $\cup_{n=1}^{\infty}I_n+x_0$ from the definition of a mesure $m^*(E)=inf(\sum_{n=1}^{\infty} |I_n|:E\subset \cup_{n=1}^{\infty}I_n)$
so we have $m^*(E+x_0)=inf(\sum_{n=1}^{\infty} |I_n+x_0|:E\subset \cup_{n=1}^{\infty}I_n+x_0)$
Looking at $\sum_{n=1}^{\infty} |I_n+x_0|\leq \sum_{n=1}^{\infty} |I_n|+|x_0|=\sum_{n=1}^{\infty} |I_n|$ taking the inf of the sum we get $m^*(E)$ as $|x_0|=0$
You were in the right path. Let us first clarify some notation. For an open interval $I=(a,b)$, we denote $|I|=b-a$.
For any interval $I=(a,b)$ and $x_0\in\mathbb R$, note that $$ I+x_0 = (a+x_0,b+x_0) \Rightarrow |I+x_0| = (b+x_0)-(a+x_0)=b-a=|I|. $$ Keep that identity in mind.
Pick $I_n$ to be open intervals such that $\displaystyle E\subset \bigcup_{n=1}^{\infty}I_n$, so we must have that $E+x_0$ is covered by $\displaystyle\bigcup_{n=1}^{\infty}I_n+x_0$. From the definition of $m^*$, $$ m^*(E+x_0)=inf\left\{\sum_{n=1}^{\infty} |J_n|:E+x_0\subset \bigcup_{n=1}^{\infty}J_n\mbox{ and $J_n$ is an open interval for any $n\in\mathbb N$}\right\}. $$ Since the intervals $I_n+x_0$ cover $E+x_0$ and $m^*(E+x_0)$ is the infimum of the sums taken on such coverings, we get that $$ m^*(E+x_0)\leq\sum_{n=1}^{\infty} |I_n+x_0|=\sum_{n=1}^{\infty} |I_n|. $$ This way we proved that for any sequence $(I_n)_{n\in\mathbb N}$ of open intervals covering $E$, $$ m^*(E+x_0)\leq\sum_{n=1}^{\infty} |I_n|, $$ then, taking the infimum on these sums we get that $$ m^*(E+x_0)\leq inf\left\{\sum_{n=1}^{\infty} |I_n|:E\subset \bigcup_{n=1}^{\infty}I_n\mbox{ and $I_n$ is an open interval for any $n\in\mathbb N$}\right\} = m^*(E). $$ This way we proved that for any $E$ and any $x_0\in\mathbb R$, $$ m^*(E+x_0)\leq m^*(E). $$ Therefore, this inequality holds if we replace above $E$ by $E+x_0$ and $x_0$ by $-x_0$, so $$ m^*(E)=m^*((E+x_0)-x_0)\leq m^*(E+x_0). $$ So $m^*(E)=m^*(E+x_0)$.