Let $\dot x=f(x), x(0)=x_0$ and initial value problem. Show that if $f\in C^1(\Bbb R)$ the solution is unique 
Let $\dot x=f(x), x(0)=x_0$ and initial value problem. Show that if $f\in C^1(\Bbb R)$ the solution is unique.

This is the problem 1.10 of the book ODE and dynamical systems of Teschl. I dont have a clue about how to show it. My work so far:
Let $\phi,\psi$ two distinct solutions to the IVP, and we will denote by $I$ it maximal and common interval of existence. Intuitive approach: WLOG exists some $t_1\in I$ (a bifurcation point) and a $\delta>0$ such that $\phi\neq\psi$ in $(t_1,t_1+\delta)$ and they are also injective (what imply that $\dot\phi\neq\dot\psi$).
From this idea (assuming that it is correct) I dont know exactly how to develop a proof by the contrapositive of the statement to be proved, that is, that under this condition $f\notin C^1(\Bbb R)$. 
Some other thoughts related: if $f$ is not continuously differentiable then there are two possible reasons: there is a point where the derivative is not defined, or there is a point where the derivative is not continuous (essential discontinuity). But I dont get any useful idea from here.

Some hints will be appreciated, thank you.
 A: This is a standard uniqueness result in the theory of ODEs.
Anyway, a direct proof can be done as follows.
Let $x(t)$ and $y(t)$ be two solutions of the IVP, defined on a common interval $[0,T]$.
Let $M := \max\{|x(t)|, |y(t)|; t\in [0,T]\}$ and let $C := \max\{|f'(z)|; |z|\leq M\}$, so that
$$
(1) \qquad
|f(x(t)) - f(y(t))| \leq C |x(t) - y(t)|,
\qquad \forall t\in [0,T].
$$
Let us consider the non-negative function
$$
z(t) := [x(t)-  y(t)]^2 e^{-2Ct},
\qquad t\in [0,T].
$$
Using (1), we have that
$$
\dot{z} = 2 e^{-2Ct} [(\dot{x} - \dot{y})(x-y) - C (x-y)^2]
= 2 e^{-2Ct} [(f({x}) - f({y}))(x-y) - C (x-y)^2]\leq 0.
$$
Since $z(0) = 0$ and $z$ is non-negative, it follows that $z(t) = 0$ for every $t\in [0,T]$, i.e. $x(t) = y(t)$ for every $t\in [0,T]$.
A: Regards @Masacroso , here is my view :
Let $x_{a} $ and $x_{b}$ be any solutions of the ODE. Now let $ X=x_{a}-x_{b}$, then
$$X'(t)=f(x_{a})-f(x_{b}), \:\:\: X''(t)=x_{a}'(t)f'(x_{a})-x_{b}'(t)f'(x_{b})  $$
and 
$$X'(0)=f(x_{0})-f(x_{0})=0, \:\:\: X''(0)=f(x_{0})f'(x_{0})-f(x_{0})f'(x_{0})=0 $$
But also $X(0)=0$ (this is crucial). You can analyze that this pattern will continue to appear for $t>0$. So we can conclude that $X(t)=0$ for $t >0$. Hence the solution for the ODE is unique ($x_{a}=x_{b}$).
Thanks.
