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!
 A: Since $\mathbb{R}$ is second countable it has a countable base of open sets  $U_k$.
For each $x \in E$ we can find some $U_{k(x)}$ such that $x \in U_{k(x)} \subset (x-\delta_x,x+\delta_x)$. Note that
$m^* (U_{k(x)} \cap E) = 0$.
Let $K = \{k(x) \}_{x \in E} \subset \mathbb{N}$, then $E = \cup_{k \in K} (E \cap U_k)$ and
$m^* E \le \sum_{k \in K} m^* (U_{k} \cap E) = 0$.
A: What you have proved is that if $E$ is at most countable, then $m^*(E)=0.$ Although this is correct, this is not what the question asks.
The set $E$ in the problem can be uncountable. For each $x\in E,$ you have the open interval $I_{x}=(x-\delta_x, x+\delta_x)$ with the property that $m^*(I_{x}\cap E)=0.$ Note that $\{I_x:x\in E\}$ is an open cover of $E.$  Choose a countable subset $D$ of $E$ such that $E\subseteq \bigcup_{x\in D}I_x$ (Why can you do that?).
Since $E\subseteq \bigcup_{x\in D}(E\cap I_{x})$. Use the fact that $D$ Is countable and $m^*$ is countable subadditive to get that $M^*(E)\le \sum\limits_{x\in D} m^*(I_x\cap E)=0.$
