Using the definition of $m$ as an outer measure, there exist $A_i=(a_i,b_i]$ such that $A\subset \cup_i A_i$ and $\sum_i (b_i-a_i) \leq m(A) + \epsilon/2$. Let $b_i' = b_i + \epsilon 2^{-i-1}$. $G:=\cup_i (a_i,b_i')$ is an open set that contains $A$ and $m(G)\leq \sum_i (b_i'-a_i) =\epsilon/2 + \sum_i (b_i-a_i) \leq m(A) + \epsilon $.
Since $m(A)<\infty$, $m(G \setminus A)=m(G) - m(A)\leq \epsilon$.
Since $m(A)<\infty$, $A$ is approcheable by a bounded set. Indeed, $m(A) = m(\cup_n (A\cap [-n,n])) = \lim_n m(A \cap [-n,n])$. There is therefore some $N$ such that $$m(A \cap [-N,N])\geq m(A)-\epsilon/2 \quad (\star)$$
Let $A' = A \cap [-N,N]$. Since $[-N,N]\setminus A'$ has finite measure, there exists some open $G'$ such that $[-N,N]\setminus A' \subset G'$ and $m(G'\setminus ( [-N,N]\setminus A'))\leq \epsilon/2$. Let us prove that the closed set $[-N,N]\setminus G'$ fits the bill.
It's easy to prove $[-N,N]\setminus G'\subset A'$. Furthermore, $$\begin{aligned} m(A'\setminus ([-N,N]\setminus G')) &= m(A'\cap ([-N,N]^c \cup G'))\\ &= m(A'\cap G') \end{aligned}$$ and
$$\begin{aligned} \epsilon/2 \geq m(G'\setminus ( [-N,N]\setminus A')) &= m((G'\cap [-N,N]^c) \cup (G'\cap A'))\\ &\geq m(G'\cap A')\end{aligned}$$
Hence $m(A'\setminus ([-N,N]\setminus G')) \leq \epsilon/2$. Let $F = [-N,N]\setminus G'$, then $$m(A') - m(F) \leq \epsilon/2 \quad (\star \star)$$
$(\star)$ and $(\star \star)$ yield $$m(A\setminus F) = m(A) - m(F) \leq (m(A') - m(F)) + \epsilon /2 \leq \epsilon $$
Finally, $F\subset A \subset G$ and $m(G\setminus F) = m(G\setminus A) + m(A\setminus F) \leq 2\epsilon$
No, your argument does not work as is, because it is not true that an arbitrary union of measurable sets is measurable (if it were, every set would be measurable since points are measurable).
What you need to do is to write $A$ as a countable union of certain $I_n$. You can achieve this for instance by considering the set
$$
A_r=\{x\in A:\ x_k\in\mathbb Q,\ k=1,\ldots,n\}.
$$
For each $x\in A_r$ there exists $I_{x,\alpha}=(x_1-\alpha,x_1+\alpha)\times\ldots\times(x_n-\alpha,x_n+\alpha) $ such that $I_{x,\alpha}\subset A$. Then, if we let $I_{x,\alpha}=\emptyset$ when $I_{x,\alpha}\not\subset A$,
$$
A=\bigcup_{x\in A_r}\bigcup_{\alpha\in\mathbb Q_+} I_{x,\alpha}
$$
is a countable union of intervals, so measurable.
Best Answer
I am not sure if there's an easier proof, or if this one is OK. But with Dumper D Garb's help I found the following:
By the dominated convergence theorem, the function $r\mapsto m(A\cap B_r(x))$ is continuous for any measurable set $A$ and any point $x\in\mathbb{R}^n$, with domain $[0,\infty]$. Therefore $g(r)=m(E\cap B_r(x))/m(B_r(x))$ is a continuous function (with domain $(0,\infty]$ since $m(B_r(x))=0$ if and only if $r=0$). Notice that for every $r$ and every $x$, $0\leq g(r)\leq 1$.
By a density theorem, almost every point of $E$ is a point of density. Since $m(E)>0$, there exists at least one such $x\in E$. Take this to be the "$x$" in the definition of $g$. Hence $g(r)\rightarrow 1$ as $r\rightarrow 0$ (this is just the definition of being a point of density). By continuity, there exists some $r>0$ such that $1/2<g(r)$; take $\alpha=g(r)$. So $m(E\cap B_r(x))=\alpha m(B_r(x))$ and by taking $B=B_r(x)$ we have proved the claim.