We start with a little Lemma:
Lemma. Let $E\subseteq \Bbb R$. If $H\supseteq E$ is a $G_\delta$ set (countable intersection of open sets) such that
$$m(H)=m^\ast(E),$$
then for every $C\subseteq\Bbb R$
$$m^\ast(H\cap C)=m^\ast(E\cap C).$$
Proof. Let $C\subseteq\Bbb R$. In the following the superscript $^c$ means complement.
$$\begin{align*}
m^\ast(H\cap C) &\leq m^\ast(H\cap C\cap E\cap C)+m^\ast((H\cap C)\setminus (E\cap C))\\
&= m^\ast(E\cap C) + m^\ast((H\cap C)\cap (E\cap C)^c)\\
&= m^\ast(E\cap C) + m^\ast(C\cap (H\setminus E))\\
&\leq m^\ast(E\cap C) + m^\ast(H\setminus E)\\
&= m^\ast(E\cap C)
\end{align*}$$
The inequality $m^\ast(H\cap C)\geq m^\ast(E\cap C)$ comes free by the monotony of the outer measure since $H\supseteq E$.
Proof of $m^\ast(E\cap (A\cup B))\geq m^\ast(E\cap A)+m^\ast(E\cap B)$.
Pick $H\supseteq E$ a $G_\delta$ set so that $m(H)=m^\ast(E)$. Then
$$\begin{align*}
m^\ast(E\cap (A\cup B)) &= m^\ast(H\cap (A\cup B)) &&\text{by the Lemma}\\
&= m(H\cap A) + m(H\cap B) &&\text{($^\ast$)}\\
&\geq m^\ast(E\cap A) + m^\ast(E\cap B) &&\text{by the monotony of the outer measure.}
\end{align*}$$
($^\ast$) because here we are dealing with measurable sets (of finite measure).
Observation. Notice that such a $G_\delta$ set $H$ always exist even if $E$ is unbounded.
Hint: Think about Bass's Proposition 4.14 (1): given any $\delta > 0$, there is an open set $G$ with $A \subset G$ and $m(G - A) < \delta$. Then recall (or prove) that every open set is a countable union of disjoint open intervals.
Here are some more details. An argument like yours has difficulties with the possibility $m(A) = \infty$, but we can reduce to the case $m(A) < \infty$ by intersecting $A$ with large bounded sets, as below.
Suppose $\epsilon \in (0,1)$ is such that $m(A \cap I) \le (1-\epsilon)m(I)$ for each interval $I$. Let $k$ be a positive integer and let $A_k = A \cap [-k,k]$. Note that $m(A_k) \le 2k < \infty$, and that for each interval $I$, we have $$m(A_k \cap I) \le m(A \cap I) \le (1-\epsilon) m(I). \tag{1}$$
Let $\alpha > 0$ be arbitrary, and using Proposition 4.14, choose an open set $G$ with $A_k \subset G$ and
$$m(G - A_k) < \alpha \epsilon. \tag{2}$$
In particular, $m(G) = m(A_k) + m(G-A_k) < 2k + \alpha \epsilon < \infty$.
Now $G$ can be written as a disjoint union of open intervals: $G = \bigcup_{n=1}^\infty I_n$. Note that $m(I_n) \le m(G) < \infty$ for each $n$. Also, $G - A_k = \bigcup_{n=1}^\infty (I_n - A_k)$ which is also a disjoint union.
Now since $I_n - A_k = I_n - (I_n \cap A_k)$, we have
$$m(I_n - A_k) = m(I_n) - m(I_n \cap A_k) \ge m(I_n) - (1-\epsilon)m(I_n) = \epsilon m(I_n) \tag{3}$$ using (1). So by countable additivity,
$$m(G - A_k) = \sum_{n=1}^\infty m(I_n - A_k) \ge \sum_{n=1}^\infty \epsilon m(I_n) = \epsilon m(G). \tag{4}$$
Combining (1) and (3), we get $\epsilon m(G) < \alpha \epsilon$, so $m(G) < \alpha$. In particular, since $A_k \subset G$, we have $m(A_k) < \alpha$. But $\alpha > 0$ was arbitrary, so we must have $m(A_k) = 0$.
Moreover, $k$ was arbitrary, so $m(A \cap [-k,k]) = 0$ for every $k$. Since $A = \bigcup_{k = 1}^\infty (A \cap [-k,k])$, by countable additivity we conclude $m(A) = 0$.
Now let's drop the assumption $m(A) < \infty$. For any $n$ and any interval $I$, we have $$m(A \cap [-n,n] \cap I) \le m(A \cap I) \le (1-\epsilon)m(I).$$
Hence $A \cap [-n,n]$ satisfies the same condition and moreover $m(A \cap [-n,n]) \le 2n < \infty$. So by the previous case, $m(A \cap [-n,n]) = 0$. Now since $A = \bigcup_{n=1}^\infty (A \cap [-n,n])$, by countable additivity $m(A) = 0$.
For an explicit example with a fat Cantor set, let's consider the example given on Wikipedia, where at stage $n \ge 1$ we remove $2^{n-1}$ intervals, each of length $2^{-2n}$. The final set $C$ has measure $1 - \sum_{n=1}^\infty 2^{n-1} \cdot 2^{-2n} = 1 - \sum_{n=1}^\infty 2^{-n-1} = \frac{1}{2}$.
Let $I_k$ be the leftmost interval that remains after stage $k$. At stage $k$ the leftmost interval that was removed was centered at $2^{-k}$ and had length $2^{-2k}$, so the leftmost interval that remains is
$$I_k = \left[0, 2^{-k} - \frac{1}{2} 2^{-2k}\right] = [0, 2^{-k}(1-2^{-k-1})].$$
At the next stage, we will remove from $I_k$ one interval of length $2^{-2(k+1)}$ from $I$, then two intervals of length $2^{-2(k+2)}$ and so on. So the total length of the intervals removed from $I_k$ is
$$\sum_{n=1}^\infty 2^{n-1} 2^{-2(k+n)} = 2^{-2k} \sum_{n=1}^\infty 2^{-n-1} = 2^{-2k} \frac{1}{2} = 2^{-2k-1}.$$
Therefore, we have
$$\frac{m(C \cap I_k)}{m(I_k)} = \frac{m(I_k) - 2^{-2k-1}}{m(I_k)} = \frac{2^{-k}(1-2^{-k-1}) - 2^{-2k-1}}{2^{-k}(1-2^{-k-1})} = \frac{(1-2^{-k-1}) - 2^{-k-1}}{(1-2^{-k-1})}$$
upon cancelling a factor of $2^{-k}$. It is clear by inspection that $$\lim_{k \to \infty} \frac{m(C \cap I_k)}{m(I_k)} = 1,$$ so given $\epsilon > 0$ you can choose $k$ so large that $m(C \cap I_k) > (1-\epsilon) m(I_k)$.
Best Answer
Let $E_n = E \cap [n\epsilon,(n+1)\epsilon)$, we have $\sum_k m E_k < \infty$.
Choose $K$ such that $\sum_{|k| >K} m E_k < \epsilon$. Let $E'=\cup_{|k|>K} E_k$.
Then $E_{-K},E_{-K+1},...,E_K, E'$ are disjoint, measurable and each have measure $\le \epsilon$.