There are similar posts, such as this one or this one, but I can't see where the right-continuity assumption is required.
Let $c:\mathbb{R}_+ \rightarrow \mathbb{R}_+$ be a non-decreasing function, right-continuous function.
Let $a:\mathbb{R_+} \rightarrow \mathbb{R_+}$ be defined by $a(u)=\inf \left\{t \in \mathbb{R_+} \,\colon \; c(t) > u \right\}$.
The claim is that $a$ is increasing and right-continuous. As for right-continuity, one should prove that for every sequence $(u_n)_{n=1}^{\infty} \subset \mathbb{R_+}$ such that $u_n \searrow u \in \mathbb{R_+} $, $ \; a(u_n) \searrow a(u)$.
So, let $(u_n)_{n=1}^{\infty}$ be such a sequence, and define, for each $n \geq 1$, $A_{u_n}=\left\{t \in \mathbb{R_+} \; \colon c(t) > u_n \right\}=c^{-1}\big(u_n, +\infty\big)$.
It is immediate to see that $(A_{u_n})_{n=1}^{\infty}$ is an increasing sequence of sets, and that for any $u \in \mathbb{R_+}$, $ \; a(u)=\inf \left\{ A_{u} \right\}$.
So, $A_{u_n} \nearrow \bigcup_{n=1}^{\infty}A_{u_n}=\bigcup_{n=1}^{\infty}c^{-1}(u_n, \infty)=c^{-1}(\bigcup_{n=1}^{\infty}(u_n, \infty))=c^{-1}\big(u, \infty \big)=\left\{t \in \mathbb{R}_+ \colon c(t) > u \right\}=A_u$. My issue is that I can't see where this list of equalities makes use of the right-continuity assumption on $c \,$:
$\bigcup_{n=1}^{\infty}A_{u_n} = \bigcup_{n=1}^{\infty}c^{-1}(u_n, \infty)$ follows by definition of $A_{u_n}$;
$\bigcup_{n=1}^{\infty}c^{-1}(u_n, \infty)=c^{-1}(\bigcup_{n=1}^{\infty}(u_n, \infty))$ follows from measurability of $c$ with respect to the Lebesgue measure on $\mathbb{R_+}$;
$c^{-1}(\bigcup_{n=1}^{\infty}(u_n, \infty))=c^{-1}(u, \infty)$ follows from the fact that $u_n \searrow u_n$.
Once the equality $\bigcup_{n=1}^{\infty} A_{u_n} = A_u$ is proven, the rest of the claim follows smoothly:
By definition, $a(u_n)=\inf\left\{ A_{u_n} \right\}$, hence \begin{align} \lim_{n \rightarrow \infty}a(u_n)=\lim_{n \rightarrow \infty} \inf A_{u_n}=\inf \left\{ \lim_{n \rightarrow \infty} A_{u_n} \right\} = \inf \left\{ \bigcup_{n=1}^{\infty}A_{u_n} \right\}=\inf \left\{ A_u \right\}=a(u) \end{align}
I am just not sure why it is possible to interchange $\inf$ and $\lim$ in
\begin{align} \lim_{n \rightarrow \infty} \inf \left\{ A_{u_n} \right\} =\inf \left\{ \lim_{n \rightarrow \infty} A_{u_n} \right\} \end{align}
but my main concern is about where the right-continuity of $c$ is required in the proof. Thanks to anyone who can help!
I don't think that right continuity of the function $c(t)$ is necessary. For $a(u)$ to be right continuous we only need for $c(t)$ to be increasing. You could see this by noting that
$$\{t: c(t) > u\} = \cup_{\epsilon > 0}\{t: c(t) > u + \epsilon\}$$
Therefore
$$\inf\{t: c(t) > u\} = \inf \cup_{\epsilon > 0}\{t: c(t) > u + \epsilon\}$$
The right hand side of this equation is given by
$$\inf \cup_{\epsilon > 0}\{t: c(t) > u + \epsilon\} = \inf_{\epsilon > 0 }\inf\{t: c(t) > u + \epsilon\} = \inf_{\epsilon > 0} a(u + \epsilon)$$
Since $a$ is increasing, $a(u+) = \lim_{\epsilon \to 0}a(u + \epsilon) = \inf_{\epsilon > 0}a(u + \epsilon)$. But we already established that $\inf_{\epsilon > 0}a(u+\epsilon) = a(u)$. Therefore $a(u+) = a(u)$, as desired.