Proving that $m^*(E) = 0.$

$\def\R{{\mathbb R}}\def\N{{\mathbb N}}$

Let $E\subseteq \R.$ Assume that for any $x\in E,$ there exists $\delta_x > 0$ so that $$m^*(E\cap(x-\delta_x,x+\delta_x))=0.$$ Prove that $m^*(E) = 0.$

$\textit{Proof.}$ Let $E$ be a finite set, say $E = \{a_1,\dots, a_k\} \subseteq \R.$ Then for every $\delta_x > 0, E\subseteq \bigcup_{i=1}^k I_i,$ where $I_i = (x_i-\delta_x,x_i+\delta_x).$ Hence $m^*(E) \le 2k\delta_x.$ Since it is true for every $\delta_x > 0, m^*(E) = 0.$

Next, suppose $E$ is a countably infinite set, say $E = \{x_i : i\in \N\}.$ Then taking $$I_n := (x_n - \frac{\delta_x}{2^{n+1}}, x_n + \frac{\delta_x}{2^{n+1}}),$$ we have $E\subseteq \bigcup_{n\in\N} I_n$ so $$m^*(E) \le \sum_{n\in \N} \ell(I_n) = \sum_{n\in\N} (\frac{\delta_x}{2^n})\le \delta_x.$$ Since this is true for all $\delta_x > 0,$ we get $m^*(E) = 0,$ and we are done.

Can I please have feedback on my proof? Thanks!