Bounded set has finite outer measure

Let $p\in E$ be arbitrary, $I_1=[p-M,p+M]$. Then $E\subseteq I_1$, hence $m^*(E)\le l(I_1)$.

If $E$ is bounded, then by definition there exists a positive finite $M$ such that $|x|\le M$ for every $x\in E.$ It follows that $E\subset [-M,M].$

The definition of $m^*(E)$ usually involves covers of $E$ by open intervals. Because of that, let's note that $E\subset (-2M,2M).$

Let's think about the definition

$$\tag 1 m^*(E) = \inf \left\{ \sum\limits_{k=1}^{\infty} l(I_k) : E \subset \bigcup\limits_{k=1}^{\infty} I_k \right\}.$$

If we abuse notation a bit and assume that finite covers and sums are allowed in $(1),$ then things are easy: We just take $I_1=(-2M,2M),$ which is a cover of $E.$ We then get $m^*(E)\le 4M.$

If we insist that $\sum\limits_{k=1}^{\infty} l(I_k)$ and $\bigcup\limits_{k=1}^{\infty} I_k$ be actual infinite series and unions, we're still fine if we are careful to stipulate that the empty set is an open interval of length $0.$ In this case we define $I_1= (-2M,2M)$ and $I_k =\emptyset$ for $k>1.$ We then see $\{I_k\}$ is a cover of $E,$ hence

$$ m^*(E) = \le \sum\limits_{k=1}^{\infty} l(I_k) = 4M+ 0+ 0+\cdots =4M.$$

Finally, if we are in a religious cult or an ancient Greek, we might insist each $I_k$ have positive length. In this case we can define $I_1 = (-2M,2M),$ and $I_k = (0,1/2^k), k>1.$ Then the open intervals $I_k$ cover $E.$ Thus

$$m^*(E) \le \sum_{k=1}^{\infty}l(I_k) = 4M + \sum_{n=2}^{\infty}\frac{1}{2^k} = 4M +\frac{1}{2}<\infty.$$