Is there an ODE so that every solution is extensible?

The following question arose while studying initial value problems ($y(t_o)=y_0, y'(t)=f(t,y(t)))$:
An solution to an IVP is a function $u:J\to \Omega$, where $J$ is an open interval and $\Omega\subseteq\mathbb{R}^n$ open which fulfills our IVP.
An Extension is a solution to our IVP $\overline{u}:\overline{J}\to\Omega$ where $\overline{J}$ is a bigger interval, $u=\overline{u}$ in $J$.
We know if $u$ is not extensible, that $u$ is either a global solution or $u(J)$ gets arbitrarily close to the boundary of $\Omega$ or $u$ has an asymptote at one of the boundary points of $J$.
Now if $f$ is locally Lipschitz continuous we know, that there exists a non extensible solution. If $f$ is only continuous however, we only know, that there exists a local solution for some J. Now do the above conditions imply a non-extensible solution for a non Lipschitz but continuous $f$?
If a solution is neither global, gets arbitrarily close to the boundary of $\Omega$ or has an asymptote, the contra position of the above Theorem states that it should be extensible. The proof provided for the existence of a non-extensible solution if $f$ is Lipschitz does not work for continuous functions and I'm wondering if there is a counter example to an analogous statement.