Let $\gamma(t)$ be the maximal integral curve with initial condition $\gamma(0) = m$, and let's suppose $b(m) < \infty$.
If we denote $\gamma'(t)$ the maximal integral curve with initial condition $\gamma'(0) = \gamma(b(m) -\epsilon/2)$ then $ \gamma''(t) = \begin{cases} \gamma(t) & \text{if } t\in (a(m), b(m) - \epsilon/2)\\ \gamma'(t - b(m) + \epsilon/2)& \text{if } t\in [b(m) - \epsilon/2, b(m) + \epsilon/2) \end{cases} $ prolongs $\gamma(t)$. That contradicts the assertion that $\gamma(t)$ is maximal.
A similar reasoning can be used to show $a(m) = -\infty$.
Edit - Some clarifications
$\gamma''$ prolongs $\gamma$ means that $\gamma''$ has an interval of existence, $(a(m), b(m) + \epsilon/2)$, that is a proper superset of the interval of existence, $(a(m), b(m))$, of $\gamma$ and the two curves are equal on the common domain, $(a(m), b(m))$.
The "uniformity" of $\epsilon$ ensures that the integral curve with initial condition $ \gamma'(0) = \gamma(b(m) - \epsilon/2) $ exists on the interval $[0, \epsilon)$. Without "uniformity" it could happen that the maximal right interval of existence of $\gamma'$ is $[0, \epsilon/2)$. In such a case, a curve $\gamma''$, constructed as above, would coincide with $\gamma$: it would not prolong $\gamma$.
The initial condition of $\gamma'$ is chosen in such a way to allow us to smoothly join $\gamma'$ to $\gamma$ in order to form a new curve, $\gamma''$, that prolongs $\gamma$.
The contradiction arises because we constructed an integral curve that prolongs a maximal integral curve, which, as such, cannot be prolonged by definition.