Consider the following differential equation \begin{equation} \begin{cases} y'(t)=f(t,y(t))\\ y(t_{0})=x_{0}\in\mathbb{R}^{n} \end{cases}\label{eq:EDO} \end{equation} where $f:[t_{0},\infty)\times\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a continuous function which is locally Lipschitz with respect to the second variable, i.e for each $b\geq 0$ there exists $L_b$ such that for all $t\geq t_0$ and $x,y\in\mathbb{R}^{n}$ such that $|x-x_0|\leq b$ and $|y-x_0|\leq b$: $$|f(x)-f(y)|\leq L_b |x-y|.$$
We have already proved that for such an equation there's a unique solution $x(t)$ defined on a maximal interval of existence $[t_{0},t_{\max}).$
Now I am asked to prove that if $t_{max}<\infty$ then ${\displaystyle \limsup_{t\to t_{\max}}|x(t)|=\infty}$. I came up with a proof which I am not sure if it is mathematically acceptable:
Assume that $t_{max}<+\infty$. If ${\displaystyle \limsup_{t\to t_{max}}|x(t)|<\infty}$ then $t\mapsto |x(t)|$ is bounded on $[t_{0},t_{\max})$ by some constant $b$. Now since $f$ is locally Lipschitz we have $$|x'(t)|=|f(t,x(t))|\leq bL_b+|f(t,0)|.$$ Since $f$ is continuous then $|x'(t)|$ is bounded on $[t_{0},t_{\max})$ which implies that $x$ is uniformly continuous on $[t_{0},t_{\max})$ so it can be extended by continuity to the interval $[t_{0},t_{\max}]$. Now if we solve the following equation
\begin{cases} y'(t)=f(t,y(t))\\ y(t_{\max})=x(t_{\max})\in\mathbb{R}^{n} \end{cases} We get a solution on $[t_{\max},t_{\max}+\varepsilon]$ that we can "stick" it to the first solution to obtain a solution on $[t_0,t_{\max}+\varepsilon]$ which thus contradicts the maximality of $t_{\max}$.