So I read parts of Rockafellar's "Convex Analysis".
When introducing "locally simplicial" sets, all the examples stated are convex sets, yet he mentions that they do not need to be convex.
I wonder whether all open sets in $\mathbb{R}^n$ are locally simplicial? I think such a fact would have been included in the reference if it were true.
My reasoning goes as follows.
A subset $S \subset \mathbb{R}^n$ is locally simplicial, if for all $x\in S$ there is a finite collection of simplices $\{S_1, \dots, S_m\}$ such that for some neighborhood $U$ of $x$ it holds $$ U \cap S = U \cap (S_1 \cup \dots \cup S_m).$$
If I assume $S$ to be open, then for every $x$ there is a ball with radius $\epsilon$ around $x$ that is still in $S$. If I place a single simplex $S_1$ that contains $x$ in its interior into that ball and take $U=S_1 \setminus \partial S_1$, I have that $U$ is a neighborhood of $x$ and it holds $$U=U\cap S = U \cap S_1.$$
This community wiki solution is intended to clear the question from the unanswered queue.
Yes, your proof is correct.