I've been stuck on this question for days. I don't even know how to begin answering the question. Any help would be greatly appreciated.
https://i.stack.imgur.com/SMZ93.png
There is a typo - the set $E$ should be such that $E\subset \mathbb{R^2}$, not $\mathbb{R}$
I'm not quite sure which facts you're allowed to assume and which you're not, but this is how I think about it...
My definition of the outer measure is $$ \theta(E) = \inf \{ \sum_{i \in \mathbb N} \mu(I_i) \ : \ E \subset \bigcup_{i \in \mathbb N} I_i \}$$ where the $I_i$'s are open rectangles and the notation $\mu(I_i)$ denotes the area of the rectangle $I_i$. The infimum is taken over all countable collections of open rectangles that cover the set $E$. Note that $\mu$ is also used as the notation for the Lebesgue measure, but there is no inconsistency since the Lebesgue measure of a rectangle is equal to its area.
By the definition of "inf", this implies that, for every $\epsilon > 0$, you can find a collection of open rectangles $I_i$ with $E \subset \bigcup_{i \in \mathbb N} I_i$, such that $$ \sum_{i \in \mathbb N} \mu(I_i) \leq \theta(E) + \epsilon $$
But $\bigcup_{i \in \mathbb N} I_i$ is a union of open rectangles, and a union of open sets is another open set! So we may as well take $$A = \bigcup_{i \in \mathbb N} I_i.$$
By countable subadditivity of the measure, $$ \mu(A) \leq \sum_{i \in \mathbb N} \mu(I_i).$$
Thus we have constructed an open set $A$ with $E\subset A$ such that $$ \mu(A) \leq \theta(E) + \epsilon .$$
This is the difficult part. I hope you can manage the rest.