The answer to the second question is yes. Trivially, every closed convex set is closed and hence in the Borel $\sigma$-algebra as the complement of an open set.
For the other direction, note that every open or closed ball is convex. Every open ball is an incrasing union of closed balls. And every open set is a countable union of open balls, since $\mathbb{R}^n$ is separable.
If one wants to restrict the problem to a compact, convex subset of $\mathbb{R}^n$, one can use the fact that a trace $\sigma$-algebra is generated by the traces of generators. That is, if $(Y,\mathcal{Y})$ is a measurable space and $\mathcal{G}\subseteq 2^Y$ such that $\mathcal{Y}=\sigma(\mathcal{G})$, and $X\subseteq Y$, then the $\sigma$-algebra $\{B\cap X:B\in\mathcal{Y}\}$ equals $\sigma\big(\{X\cap G:G\in\mathcal{G}\}\big)$.