I wrote up some proofs for some claims and wish to have someone critic/provide a really straightforward proof of one that I can model after for at least the 'application' of proving a set is/not R-open/closed.
Let R = { V $\subset$ $\mathbb{R}$ $\vert$ if x $\in$ V, then there exists a, b $\in$ $\mathbb{R}$ a < b such that x $\in$ (a, b] $\subset$ V }
Claim. A = [-1, 0) is not R-closed.
Proof. Consider $\mathbb{R}$\A. If x $\in$ $\mathbb{R}$\A, then x $\in$ $\mathbb{R}$ and x $\not$$\in$ [-1, 0). Then x $\in$ (-$\infty$, 1) $\cup$ [0, $\infty$). If x $\in$ (-$\infty$, 1), then x $\in$ [x, 1) $\subset$ (-$\infty$, 1). Else if x $\in$ [0, $\infty$), then 0 $\in$ [0, $\infty$). For any a > 0, 0 $\not$$\in$ (a, $\infty$) $\subset$ V. Hence $\mathbb{R}$\A $\not$$\subset$ V $\in$ R implies that $\mathbb{R}$\A $\not$$\in$ R. Therefore $\mathbb{R}$\A is not R-open and hence A is not R-closed.
Claim. [-1, 0) = A is not R-open.
Proof. Let x $\in$ A so that x $\in$ [-1,0). Then -1 $\in$ [-1,0). For any a > -1, -1 $\not$$\in$ (a, 0] $\subset$ V $\in$ R. Hence A $\not$$\subset$ V $\in$ R implies that A $\not$$\in$ R.
Claim. C = (-$\infty$, - 1] $\cup$ (0, $\infty$) is R-open.
Proof. Let x $\in$ (-$\infty$, - 1] $\cup$ (0, $\infty$). Then if x $\in$ (-$\infty$, -1], then x $\in$ (x - 1, -1] with x - 1, -1 $\in$ $\mathbb{R}$, x $\in$ (x - 1, -1] $\subset$ (-$\infty$, -1] $\subset$ V. Else if x $\in$ (0, $\infty$), then x $\in$ (0, x + 1] with 0, x + 1 $\in$ $\mathbb{R}$, then x $\in$ (0, x + 1]$\subset$ (0, $\infty$) $\subset$ V. Hence, x $\in$ (-$\infty$, -1] $\cup$ (0, $\infty$) $\subset$ V $\in$ R.
PS. I'm very appreciated of this site and that fact I get comments/replies. The feedback helps so much.