Suppose $F:[0, \infty) \rightarrow [0, \infty)$ is nondecreasing and right-continuous with jump discontinuities at each point in the set $\{x_1, x_2, \ldots, x_k\}$.
Define $c^*(y)=\max\{ x \in \{x_1, \ldots, x_k\} \mid F(x) \le g(y)\}$ for $y >0$ and an increasing function $g$ defined on $[0, \infty)$. I want to prove that $\inf\{y \in [0, \infty) \mid x_0 \le c^*(y)\}$ is $\inf\{y \in [0, \infty) \mid F(x_0) \le g(y)\}$ for some $x_0 \in \{x_1, \ldots, x_k\}$.
Sorry, but I think your assumption is incorrect. I think it would be $F(x_{n+1})$ such that $x_n<x_0\le x_{n+1}$ where $x_n,x_{n+1}\in \{x_1,x_2,\dots,x_k\}$. Also for $x_0> x_k$, the infimum, wouldn't exist as there would be no $c(y)\ge x_0$.
Here's why. Let $J:= \{x_1,x_2,x_3,\dots, x_k\}$. Let $I(x_0):= \inf\{y\in [0,\infty):x_0<c(y) \}$. Now let's look at the three cases if $x_0 \in J$, $x_n<x_0<x_{n+1}: x_n,x_{n+1}\in J$, and $x_0 > x_k$.
If $x_0 \in J$ then indeed $I(x_0)=F(x_0)$ since $c(F(x_0))=x_0$.
If $x_n<x_0<x_{n+1}: x_n,x_{n+1}\in J$. Then $C(F(x_0))=x_n$. Since $x_n<x_0$, it implies that $F(x_0)$ is a lower bound. However since $x_{n+1}>x_0$ it is possible for there to exist a $y$ such that $c(y_0)=x_{n+1}$, namely $y_0=F(x_{n+1})$. It is sufficient to now prove that any $y< y_0$ is a lower bound. Note that $c(y)=x_n$ if $y<y_0$, and thus a lower bound.
If $x_0 > x_k$. Since $c(x_0)=x_k$ this implies that $\{y\in [0,\infty): x_0<c(y)\}=\{\}$, and thus unbounded.