Proof that Quantile Function characterizes Probability Distribution The quantile function is defined as $Q(u)= \inf \{x: F(x) \geq u\}$. 
It is well known the distribution function characterizes the probability distribution in the following sense
Theorem Let $X_{1}$ and $X_{2}$ be two real valued random variables with distribution functions $F_{1}$ and $F_{2}$ respectively. If $F_{1}(x)=F_{2}(x)$, $\forall x\in \mathbb{R}$ then $X_{1}$ and $X_{2}$ have the same probability distribution. 
I want to prove that the quantile function also characterizes the probability distribution, this fact is stated as a corallary of the above theorem in this book (see Corallary 1.2 on p19):
Corallary: Let $X_{1}$ and $X_{2}$ be two real valued random variables with quantile functions $Q_{1}$ and $Q_{2}$ respectively. If $Q_{1}(u)=Q_{2}(u)$, $\forall u\in\left(0,1\right)$ then $X_{1}$ and $X_{2}$ have the same probability distribution. 
The proof in the book is based on the following facts 
Fact i: $Q(F(x)) \leq x$.
Fact ii: $F( Q(u) ) \geq u$.
Fact iii: $Q(u) \leq x$ iff $u \leq F(x)$.
Fact iv: $Q(u)$ is nondecreasing.
But I think it is wrong. Assuming $F_{1}(x_0) < F_{2}(x_0)$ for some fixed $x_0$ the author sets out to prove that this leads to a contradiction. Using facts (i) and (iv) he shows $Q_{2}(F_1(x_0)) < Q_{2}(F_2(x_0)) \leq x_0$. Then he applies fact (iii) to obtain $F_{1}(x_0) \leq F_{2}(x_0)$. The author claims that this is a contradiction. But clearly its not and the argument proves nothing. 
Am I missing something here? Does anyone know the correct proof to the corallary? 

Changed since version discussed in first five comments:
The key line  in the proof of Corollary 1.2 in Severini's Elements of Distribution Theory book is 

Hence by part (iii) of Theorem 1.8, $F_2(x_0) \ge F_1(x_0)$ so that $F_1(x_0) \lt F_2(x_0)$ is impossible.

As you say,  $F_1(x_0) \lt F_2(x_0)$ in fact implies $F_1(x_0) \le F_2(x_0)$, i.e. $F_2(x_0) \ge F_1(x_0)$, so this is not a contradiction.  
