I want to prove the following statement:
Let F fulfill the properties of a cumulative distribution function. Define $$X^-(\omega) = \inf \{z \in \mathbb{R}: F(z) \geq \omega\} \quad X^+(\omega) =\inf \{z \in \mathbb{R}: F(z) > \omega\} $$ Then $X^- = \sup \{y \in \mathbb{R}: F(y) < \omega\} = X_-$ and $X^+ = \sup \{y \in \mathbb{R}: F(y) \leq \omega\} = X_+$.
Proof. Note that if $\exists x_0: F(x_0) = \omega$. Then $X^-(\omega) = X^+(\omega)$. And obviously $X^- = X_+ = X_-$. So assume that $\not\exists x_0: F(x_0) = \omega.$ Show that $X_+ = X^+$ since the other equality follows by the same argumentation.
It holds that $X_+ \leq X^+$, since for every $y \in \{y \in \mathbb{R}: F(y) \leq \omega\}$ it holds that $F(y) \leq \omega$ thus by monotonicity if $F(z) > \omega$ this implies that $y < z$. Applying supremum and infimum we get that $\sup_{y} y \leq \inf_{z} z$. Let $z_0 = X^+(\omega)$ and $y_0 = X_+(\omega)$. Suppose that $y_0 < z_0$ (this implies that $F(y_0) < \omega$). By the monotonicity we obtain $$\forall y \in U^+(y_0): F(y) > \omega, \tag{$*$}$$ since if $F(y) \leq \omega$ we get that $y \in \{y \in \mathbb{R}: F(y) \leq \omega\}$. Yet $y > y_0$ thus $y_0$ would not be a supremum. In fact, it would not be an upper bound.
Due to the fact that F is continuous from the right we get that $$\forall \varepsilon > 0: \exists U^+(y_0): \exists y \in U^+(y_0): F(y) - F(y_0) \leq \varepsilon \tag{$**$}$$ If we rewrite $(**)$ we obtain that $$F(y) \leq F(y_0) + \varepsilon < \omega + \varepsilon, \quad \forall \varepsilon > 0.$$ This is because by assumption $y_0 < z_0$ and since $z_0$ is the infimum such that $F(z) > \omega$ (we assumed that $\omega$ will not be obtained by any valued) $F(y_0)$ has to be smaller than $\omega$. Note that this implies that $F(y) \leq \omega$ ($< \omega$, since not obtained), yet this is a contradiction to $(*)$. Thus $F(y_0) > \omega$, yet this implies that $$\omega < F(y_0) \leq F(z_0)$$ since $z_0$ is the infimum $y_0 = z_0$.
Note that we did not include the case that $\omega = 0$ or $\omega = 1$ in this case we get for $X^+$ and $X_+$ either $-\infty$ or $+\infty$ respectively by the computation rules for the infimum or supremum.
