Consider
$\dot{x}=x^{3}, x \in \mathbb{R}, x(0)=1$. T his is solved by $x(t)=$ $1 / \sqrt{1-2 t}$, so $x \rightarrow \infty$ when $t \rightarrow \frac{1}{2}$.
I have the following theorem for uniqueness 
My task is to explain why there is no contradiction.
My reasoning: Clearly the conditions for the uniquness theorem hold. What is happening is that we get a small neighbourhood about $t_0=0$ on which the solution exists and is unique, say $(-\delta_1,\delta_1)$. Now I can apply the theorem again but this time with $t_0 = \delta_1 / 2$. I can continue this process to get a unique, continuous solution onto $(0, \tau)$, where $\tau$ is the supremum of the (\delta_i) in this process. What is happening above is that this supremum is $1/2$, and so we can't use the uniqueness theorem at $1/2$ itself.
Question: Is my reasoning correct?
Edit: Additional Context (1) will be the ode mentioned in the first definition bellow. All other terms should be bellow too.
