The problem I've been struggling is exemplified here:
Is local Lipschitz continuity sufficient for an ODE to have a unique solution?
The problem that they solve is
\begin{align*} x'(t) = x^2,\quad x(0) = x_0 \end{align*}
which in the other thread is said to have a unique solution given by
\begin{align*} x(t) = \frac{1}{x_0^{-1}-t}. \end{align*}
This function is not continuous at $t = x_0^{-1}$ bu they claim in the other thread that it is unique in $(-\infty,x_0^{-1})$, supposing $x_0>0$. Since $f(x,t) = x^2$ the function is only locally Lipschitz and therefore we are only guaranteed a unique solution locally? How can one in general be sure that it does not split into separate branches outside of some small interval?
Local uniqueness already garantuees uniqueness on the whole intervall of existence. Let me give a short proof:
Let $u_1$ and $u_2$ solve the same initial value problem. Let us also assume there exists a $t\in\mathbb{R}$ such that for every $\varepsilon>0$ we find an $\tilde{t}\in(t,t+\varepsilon)$ such that $u_1(\tilde{t})\neq u_2(\tilde{t})$ and $u_1(t)=u_2(t)$. This means $u_1$ and $u_2$ are equal up to $t$, but after that $t$, they may differ. Local uniqueness now gives us a small open neighborhood of $t$ in which $u_1$ equals $u_2$. Hence we find an $\tilde{\varepsilon}>0$ such that for every $x\in(t,t+\tilde{\varepsilon})$ we have $u_1(x)=u_2(x)$. If you choose you $\varepsilon$ now to be smaller than $\tilde{\varepsilon}$ you obtain a contradiction.
Your example is also unique up to the point of existence. Please note, that your $x(t)$ tends to $\infty$ if $t\uparrow x_0$. Hence your solution cannot be extended further and therefore it cannot branch. (See also https://en.wikipedia.org/wiki/Ordinary_differential_equation#Global_uniqueness_and_maximum_domain_of_solution)