Suppose $S\subset\mathbb R$ such that $S\cap(x,\infty)\neq\phi$ for every $x\in\mathbb R$ and suppose that $g:S\to\mathbb R$ is a non-decreasing bounded function. Define $G:\mathbb R\to\mathbb R$ as: $$G(x)=\inf_{y\in S\cap(x,\infty)}g(y)$$for any $x\in\mathbb R$.
Is the function $G$ right continuous?
I observe that $G$ is non-decreasing. Somehow I believe that it is not true that $G$ is right continuous. But I cannot really prove it. I suppose the reason why I believe this is that to say that $G$ is right continuous, I would need $g$ to be right continuous, which is not given. Again, I may be wrong.
Well, note that for every decreasing sequence $x_n \to x$ you have $\bigcup_n (x_n,\infty) = (x,\infty)$. This implies $\lim_{n \to \infty} \inf_{S\cap(x_n,\infty)} g =\inf_n \inf_{S\cap(x_n,\infty)} g = \inf_{S\cap(x,\infty)}g$. Since you have in general $\inf \bigcup_{i\in I} A_i = \inf_{i\in I} \inf A_i$