I hope you can help me with the following, please
Let $X$ be a random variable with c.d.f. $F$. For each $p\in (0,1)$. Define $F^{-1}(p)$ to be the smallest value $x$ such that $F(x)\geq p$. I want to prove the following:
Let $x_0=\lim_{\{p\to 0, p>0\}}\ F^{-1}(p)$. Prove that $x_0$ is the greatest lower bound of the set of numbers $c$ such that $P(X\leq c)>0$.
Usually, when we prove that something is the greatest lower bound of a set, we first demonstrate that it is a lower bound and then that it is the greatest. However, I would like to know if there is a way to perform this proof by contradiction. I have seen the proof attached below, and it did not seem intuitive to me, especially I do not understand what is in bold. I greatly appreciate your help.
$\textbf{The proof}:$
Let $x_1=\dfrac{x+x_0}{2}$, note that $x_0<x_1<x.$ Because $F^{-1}(p)$ is a non-decreasing function, and considering that $x_0$ is essentially the right limit of $F^{-1}(p)$ as $p$ tends to $0$, $\textbf{there exists some $p>0$ such that $F^{-1}(p)<x_1$}$, which implies that $p\leq F(x_1)$ and thus $F(x_1)>0$, so $x_1\in C$ and $x$ is not a lower bound of $C$.
Now, let's see that $x_0$ is a lower bound of $C$. For any $x\in C$, we need to prove that $x_0\leq x$. Since $F^{-1}(p)$ is non-decreasing, we have $\lim_{p\to 0, p>0}\ F^{-1}(p)\leq F^{-1}(q)$ for all $q>0$. Thus, $x_0\leq F^{-1}(q)$ for all $q>0$. Since $x\in C$, we have $F(x)>0$. Let's define $q=F(x)$, then $x_0\leq F^{-1}(q)\leq x.$