Let $E\subseteq \mathbb{R}$ be a subset, we want to show that there exists a Borel set $E\subseteq B\subseteq \mathbb{R}$ such that each Lebesgue measurable subset $H\subseteq (B\setminus E)$ has measure zero.
Here's my proof attempt:
Let $E\subseteq \mathbb{R}$. We have $$m^*(E) = \inf\left\{\sum_k(b_k-a_k):E\subseteq \bigcup_k(a_k,b_k]\right\}$$ where $m^*$ denotes the Lebesgue outer measure. Hence, for each $n\in \mathbb{N}$ we can get a sequence of sets $\left\{(a_k^{(n)},b_k^{(n)}]\right\}_k$ such that $$E\subseteq \bigcup_k \left(a_k^{(n)},b_k^{(n)}\right]\text{ and }\sum_k(b_k^{(n)}-a_k^{(n)})-m^*(E) < \frac1n.$$ Now, set $B_n = \bigcup_k\left(a_k^{(n)},b_k^{(n)}\right]$ and $B=\bigcap_n B_n$. So, $B$ is a Borel set and $E\subseteq B$. It remains to show that, for all $H\subseteq B\setminus E$, where $H$ is Lebesgue measurable, we have $m(H) = 0$. We first define $$A=\bigcap_{\substack{S\in\mathcal{L}\\ E\subseteq S}}S\text{ where }\mathcal{L}\text{ denotes the set of Lebesgue measurable sets}.$$ Then, we have $A\in \mathcal{L}$ and $E\subseteq A\subseteq B$.
Claim: Suppose that $H\subseteq B\setminus E$ is a Lebesgue measurable set, then $H\subseteq B\setminus A$.
Assuming the above claim, we can complete the proof. It suffices to show that $m(B\setminus A) = 0$. Notice that, for all $n\in\mathbb{N}$, we have $$m(B\setminus A) \leq m(B_n\setminus A) = m(B_n)-m(A) \leq \sum_k(b_k^{(n)}-a_k^{(n)})-m^*(E) < \frac1n$$ if $m(B_n)<\infty$ (this allows us to make the subtraction in the complement step). However, I'm not sure how to prove the infinite case and how to rigorously prove the claim (even though it seems intuitive to me). Please let me know if I'm on the right track / provide some hints. Thanks!
Upon further attempts on my own and some suggestions from the comments, here is my improved solution to this problem (note that the first part regarding the construction of $B$ is copied from the question itself):
First, supposed that $E\subseteq \mathbb{R}$ is bounded. We have $$m^*(E) = \inf\left\{\sum_k(b_k-a_k):E\subseteq \bigcup_k(a_k,b_k]\right\}$$ where $m^*$ denotes the Lebesgue outer measure. Hence, for each $n\in \mathbb{N}$ we can get a sequence of sets $\left\{(a_k^{(n)},b_k^{(n)}]\right\}_k$ such that $$E\subseteq \bigcup_k \left(a_k^{(n)},b_k^{(n)}\right]\text{ and }\sum_k(b_k^{(n)}-a_k^{(n)})-m^*(E) < \frac1n.$$ Now, set $B_n = \bigcup_k\left(a_k^{(n)},b_k^{(n)}\right]$ and $B=\bigcap_n B_n$. So, $B$ is a Borel set and $E\subseteq B$. It remains to show that, for all $H\subseteq B\setminus E$, where $H$ is Lebesgue measurable, we have $m(H) = 0$.
Let $H\subseteq B\setminus E$ and set $A = B\setminus H$, then we have $E\subseteq A$ and $H = B\setminus A$ where $A$ is Lebesgue measurable. We then have the following (for all $n\in\mathbb{N}$): $$m(H) = m(B\setminus A) = m(B)-m(A) \leq m^*(B_n)-m^*(E) < \frac1n.$$ Hence, we have $m(H)=0$.
Now, suppose that $E$ is unbounded and we write $$E = \bigcup_j E_j\text{ where } E_j = [-j,j]\cap E.$$ Since $E_j$ is bounded for each $j\in\mathbb{N}$, we then have, for all $j$, there exists some $B_j$ such that $E_j\subseteq B_j$ and $m(H_j)=0$ for all Lebesgue measurable $H_j\subseteq B_j\setminus E_j$. Now, let $B = \bigcup_j B_j$ and consider some Lebesgue measurable $H\subseteq B\setminus E$. Now, for each $j$, consider $H_j = B_j\cap H$ and notice that $H_j\subseteq B_j\setminus E_j$ for each $j$ is Lebesgue measurable and that $H = \bigcup_j H_j$. Hence, we have $$m(H) \leq \sum_j m(H_j) = 0\implies m(H) = 0.$$ Thus, our proof is complete.