Proving that $m((-\infty, x] \cap K) \setminus m((-\infty , y] \cap K) = m((] \cap K),$ where $K$ is a compact set.
My idea:
we can use excision property, which states that: If $A$ is a measurable set of finite outer measure that is contained in $B,$ then $$m^{*} (B \setminus A) = m^{*}(B) \setminus m^{*}(A)$$.
And because every interval is measurable and every closed set of $\mathbb{R}$(in our case $K$ as it is compact) is the intersection of a countable family of open sets and every open set is the disjoint union of a countable collection of open intervals. this can change the outer measure in the above equality into a measure.i.e.$$m (B \setminus A) = m(B) \setminus m(A)$$.
And then using that the set intersection is distributed over set difference from the left will complete the proof.
Is my proof correct?
Your idea is correct.
Let $x<y$ without loss of generality.
Indeed by measurability $$m((-\infty, y] \cap K)- m((-\infty , x] \cap K)$$ $$=m(((-\infty, y] \cap K) \setminus (K \cap (-\infty , x])) $$ $$=m(((-\infty, y] \cap K) \cap (K^c \cup (x,+\infty)))$$ $$=m(K \cap (x,y])$$