Consider the initial value problem on $[0,\infty)$: $$x'(t)=f(t, x(t)) \qquad x(0)=0,\label{1}\tag{$*$}$$ where $f:(0,\infty)\times\mathbb R\to\mathbb R$ is a continuous function. Assume that for every $a>0$, there exists a $C^1$ solution $x(t)$ to the initial value problem \eqref{1} on $[0,a)$ with $x\in C([0,a))\cap C^1((0,a))$ (the solution is not necessarily unique). Can we conclude that there is a solution to the initial value problem \eqref{1} on the whole $[0,\infty)$?
This problem comes from one of my friends.
Attempt. I think the answer is “no”. To find a counter-example, I tried to construct an $f$ such that the initial value problem $(*)$ has a solution $x(t)$ on $[0,T)$ with the property $$x(t)\sim \frac1{T-t}, \qquad t\sim T-$$ for all $T>0$; and such that $(*)$ doesn't have global solutions on the whole $[0,\infty)$. However, the function $\frac1{T-t}$ doesn't take the value $0$ at $t=0$, so I considered instead $$x(t)=\frac{\eta(t)}{T-t}, \qquad t\in[0,T)\tag{1}$$ for some good function $\eta(t)$ with $\eta(0)=0$. Now the function $(1)$ satisfies $$x'(t)=\frac{\eta(t)}{(T-t)^2}+\frac{\eta'(t)}{T-t},\qquad t\in[0,T),$$ which impiles that $x$ is a solution to the following initial value problem on $[0, T)$: $$x'(t)=\frac{1}{\eta(t)}(x(t))^2+\frac{\eta'(t)}{\eta(t)}x(t),\qquad x(0)=0. \tag{2}$$ However, the initial value problem $(2)$ has a trivial global solution $x\equiv0$ and many non-trivial global solutions $x(t)=\frac{\eta(t)}{C-t}$ on $t\in[0,\infty)$ for $C<0$. Therefore, this method doesn't work.
Any help would be appreciated!
Note. This problem has been cross-posted in MO and has been answered by Saúl RM.
Nevertheless, more examples or comments are welcome!