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.
Another nice way of looking at this problem is to go through an "inductive" definition of the Borel subsets of $\mathbf{R}^n$. So, for ordinals $\alpha<\omega_1$ we inductively define the Borel hierarchy as follows:
Let $\mathbf{\Sigma}^0_1$ denote the collection of open subsets of $\mathbf{R}^n$.
Let $\mathbf{\Pi}^0_\alpha$ denote the collection of relative complements of elements of $\mathbf{\Sigma}^0_\alpha$.
Let $\mathbf{\Sigma}^0_\alpha$ denote the collection of $\bigcup_{n\in\mathbf{N}} X_n$ where each $X_n$ is in some $\mathbf{\Pi}^0_{\beta_n}$ with $\beta_n<\alpha$.
Recalling that the Borel subsets of $\mathbf{R}^n$ is smallest $\sigma$-algebra containing the open sets, we see that what we have done here is "stratify" the Borel sets. In particular, one can show that the Borel subsets of $\mathbf{R}^n$ is the same as the collection $\bigcup_{\alpha<\omega_1}\mathbf{\Sigma}^0_\alpha$. So now, we can in fact show that if $f:\mathbf{R}^m\rightarrow\mathbf{R}^n$ is continuous, then the levels of the Borel hierarchy are preserved under $f$. In particular, Borel sets are preserved by continuous preimages. Also note that, if we show $\mathbf{\Sigma}^0_\alpha$ sets are preserved under continuous preimages, then, since preimages are well behaved under complements, $\mathbf{\Pi}^0_\alpha$ sets are preserved under continuous preimages. So, it suffices to show that the $\mathbf{\Sigma}^0_\alpha$ sets are preserved under continuous preimages. We proceed by induction.
We know that the open sets, or $\mathbf{\Sigma}^0_1$ sets are preserved under continuous preimages, so assume that this is the case for all $\beta<\alpha$. Then, as we noted above, this holds for $\mathbf{\Pi}^0_\beta$ sets for all $\beta<\alpha$. So, let $A\in\mathbf{\Sigma}^0_\alpha$, then we see that $A=\bigcup_{n\in\mathbf{N}}X_n$ where each $X_n$ is in some $\mathbf{\Pi}^0_{\beta_n}$ where $\beta_n<\alpha$. Then, we see that $f^{-1}(A)=f^{-1}(\bigcup_{n\in\mathbf{N}}X_n)=\bigcup_{n\in\mathbf{N}}f^{-1}(X_n)$. But each $f^{-1}(X_n)$ is $\mathbf{\Pi}^0_{\beta_n}$ where $\beta_n<\alpha$, by our hypothesis. Thus, $f^{-1}(A)\in\mathbf{\Sigma}^0_\alpha$.
So, to answer your question (kind of), there is a nice way of expressing your original idea.
Best Answer
Looks fine to me. You should get full marks for that.