If $E \cap K$ is measurable for every compact set $K$, then so is $E$.
Measurable space is $\bf{R}^n$ with Lesbegue measure.
My idea is to generate $E$ through a sequence of $E \cap K_n$ and take its union and argue that $E$ is the countable union of measurable sets, so it is measurable.
My book has a theorem that states
If $m(E) < \infty$, then there is a $K$ compact set with $K \subset E$ and $m(E-K)<\epsilon.$
The technique sets $K_n = \bar{B_n(0)} \cap F$ with $F_{closed} \subset E$ and argues that $E - K_n$ decreases.
Is the solution here the same? Except, I use the fact that $F \subset E\cap K \subset K$?
My idea (continued): I thought of iterating $E \cap B_n(0)$, but I can't justify why $K$ is allowed to be take the closed ball.
I think that the use of the set $F$ is questionable, you haven't shown how this is constructed. On the other hand you probably don't need that set at all.
The only place where you use the fact that's the space is $\mathbb R^n$ is where you construct the sequence $K_n$. The proof would work for any space where there is a sequence of compact sets whose union is the entire space. We can by the existence of such a sequence guarantee a growing sequence of compact sets (ie $K_j \subseteq K_k$ if $j\le k$):
With such a set we then have that $E\cap K_j$ is measurable and $m(E\cap K_j)$ is non-decreasing. The measurability of $E$ follows as $E = \bigcup E\cap K_k$.
The case with infinite measure may have to be handled separately. The definition then normally requires that it will be locally measurable in order to be measurable and the result follows in that case anyway.