Consider the vector field $F:\mathbb{R}^2 \to \mathbb{R}^2$, defined by $F(x,y) = \begin{pmatrix}y \\x^2\end{pmatrix}$.
Can it be shown that $F$ is not complete?
Equivalently: Is there an explicit integral curve $\varphi:I \to \mathbb{R}^2$ with $I \subseteq \mathbb{R}$ satisfying $\dot{\varphi}(t)=F(\varphi(t))$, which is not defined on all of $\mathbb{R}$?
Let $\varphi(t)=(x(t),y(t))$; then $\dot{\varphi}(t)=F(\varphi(t))$ is equivalent to $(\dot{x},\dot{y})=(y,x^2)$, which implies $\ddot{x}=\dot{y}=x^2$. Integrating it, we obtain
$\frac{1}{2}\dot{x}^2=\frac{1}{3}(x^3+C)$,
where $C=\frac{1}{2}y_0^2-\frac{1}{3}x_0^3$. Integrating again, we obtain
$t=\int_{x_0}^x\frac{du}{\sqrt{\frac{1}{3}(u^3+C)}}$,
where we assumed that $\dot{x}(0)=y_0>0$ in order to select the positive square root of $\dot{x}^2$, and also to ensure the convergence of the integral at the lower limit of integration. Since the integral also converges in the limit $x\to\infty$, it follows that $x(t)$ is defined only for $t\in[0,T)$, where
$T=\int_{x_0}^{\infty}\frac{du}{\sqrt{\frac{1}{3}(u^3+C)}}<\infty$.
This shows that $F$ is not complete. $\square$