Let $D$ be a Jordan-measurable subset of $\mathbb{R}^n$ and $S$ be a null set (By a null set I mean a set whose Lebesgue (outer) measure equals 0). In general $D\setminus S$ would not be Jordan-measurable, as the example $S=\mathbb{Q}$ shows. However, my textbook says that if we assume further that the set $D\setminus S$ is open, then $D\setminus S$ must be Jordan-measurable. The book does not give a detailed explanation so I tried to figure this out by myself. I would like to check if my argument is valid, and since my argument is somewhat lengthy, if shorter explanation is possible.
We may assume $S\subset D$. Since $S$ is a null set, its interior must be empty and we have $\overline{S}=\text{Int}(S)\cup\text{Bd}(S)=\text{Bd}(S)$. Since $D\setminus S$ is open and disjoint from $S$ we have $(D\setminus S)\subset\text{Ext}(S)$, in other words, $(D\setminus S)\cap\overline{S}=\varnothing$.
Let $p\in\overline{S}$ and assume that $p\notin S$: If $p\in\text{Int}(D)$ then $p$ becomes a member of $D\setminus S$ but this is impossible since $(D\setminus S)\cap\overline{S}=\varnothing$. Our assumption $S\subset D$ implies that $p\in\overline{S}\subset\overline{D}=\text{Int}(D)\cup\text{Bd}(D)$, so we have $p\in\text{Bd}(D)$.
Thus if $p\in\overline{S}$, then either $p\in S$ or $p\in\text{Bd}(D)$: we have shown that $\overline{S}\subset S\cup\text{Bd}(D)$. This implies $S\cup\text{Bd}(D)=\overline{S}\cup\text{Bd}(D)$, and this last expression is a union of two compact sets. On the other hand, the expression $S\cup\text{Bd}(D)$ is a union of two null sets (since $D$ is Jordan-measurable, $\text{Bd}(D)$ is a null set), so this set is also a null set.
Finally since $\text{Bd}(D\setminus S)\subset\text{Bd}(D)\cup\text{Bd}(S)=\text{Bd}(D)\cup\overline{S}$, the set $\text{Bd}(D\setminus S)$ is a subset of a null set, hence is a null set on its own. Hence $D\setminus S$ is Jordan-measurable.
Is the above argument valid?
Your proof looks reasonable.
I this it is a little easier to read (and shorter) if you deal with the sets directly.
Since $D \setminus S \subset D$ we have $\overline{D \setminus S} \subset \overline{D}$. Since $D \setminus S$ is open, $(D \setminus S)^c$ is closed and so $\partial (D \setminus S) \subset \overline{D} \cap (D^c \cup S) \subset \overline{D} \cap (\overline{D^c} \cup S)\subset \partial D \cup S$.