Application of Gronwall Inequality to existence of solutions 
Consider the $N$-dimensional autonomous system of ODEs
  $$\dot{x}= f(x),$$
  where $f(x)$ is defined for any $x \in \mathbb{R}^N$, and satisfies $||f(x)|| \leq \alpha||x||$, where $\alpha$ is a positive scalar constant, and the norm $||x||$ is the usual quadratic norm (the sum of squared components of a vector under the square root). Using Gronwall’s inequality, show that the solution emerging from any point $x_0\in\mathbb{R}^N$ exists for any finite time.

Here is my proposed solution. 
We can first write $f(x)$ as an integral equation,
$$x(t) = x_0 + \int_{t_0}^{t} f(x(s)) ds$$
where the integration constant is chosen such that $x(t_0)=x_0$. WLOG, assume that $t_0=0$. Then,
\begin{equation}
\begin{split}
||x(t)|| & = ||x_0 + \int_{0}^{t} f(x(s)) ds|| \\
 & \leq ||x_0|| + ||\int_{0}^{t} f(x(s)) ds|| \\
 & \leq ||x_0|| + \alpha\int_{0}^{t} ||x(s)|| ds 
\end{split}
\end{equation}
Therefore, by the integral form of Gronwall's inequality, we see that
\begin{equation}
\begin{split}
||x(t)|| & \leq ||x_0|| + \alpha\int_{0}^{t} ||x(s)|| ds \\
 & \leq  ||x_0||e^{\alpha(t)}
\end{split}
\end{equation}
So, if we let $M = ||x_0||$, then $||x(t)||\leq{{M}e^{\alpha(t)}}$. Therefore, the solution is uniformly bounded on $[0,t]$ for $t>0$.
As $t>0$ was arbitrary, the solution is defined for all positive values of $t$.
We can then analyze what happens for negative values of $t$ by reversing time and applying the same argument to $[-t,0]$.
Once again assume that $t_0=0$. Then,
$$x(t) = x_0 + \int_{-t}^{0} f(x(s)) ds$$
Therefore,
\begin{equation}
\begin{split}
||x(t)|| & = ||x_0 + \int_{-t}^{0} f(x(s)) ds|| \\
 & \leq ||x_0|| + ||\int_{-t}^{0} f(x(s)) ds|| \\
 & \leq ||x_0|| + \alpha\int_{-t}^{0} ||x(s)|| ds \\
 & \leq ||x_0||e^{\alpha(0+t)} \\
 & = {M}e^{\alpha(t)}
\end{split}
\end{equation}
So, the solution is uniformly bounded on $[-t,0]$ for $t<0$.
Combining these two bounds, we see that the solution emerging from any point $x_0\in\mathbb{R}^N$ exists for any finite time.
Is this approach correct? Please let me know if there are any better alternatives.
 A: As explained in the comments, the proposed answer only shows that the solution is bounded between some arbitrary interval $[t_1,t_2]$ where $t_1,t_2,\in\mathbb{R}$. We also need to show that we can extend the solution to any interval of finite length.
To do this, consider Lemma $2.14$ on page $52$ of Teschl.
$\textbf{Lemma 2.14:}$ Let $\phi(t)$ be a solution of $(2.10)$ defined on the interval $(t_-,t_+)$. Then there exists an extension to the interval $(t_-,t_+ + \epsilon)$ for some $\epsilon > 0$ if and only if there exists a sequence $t_m\in(t_-,t_+)$ such that
$$\lim_{m\to\infty}(t_m,\phi(t_m))=(t_+,y)\in{U}. $$
The analogous statement holds for an extension to $(t_- - \epsilon,t_+).$ 
As $||x(t)||\leq{{M}e^{\alpha(t)}}$, it is clear that $x$ lies in a compact ball. Therefore, by Lemma $2.14$ (and the Bolzano–Weierstrass theorem), we can extend the solution to any interval of finite length.
An alternative argument would be using Corollary $2.16$. I don't like the way Corollary $2.16$ is phrased and have decided to directly apply Lemma $2.14$ instead.
