Maximal Solution (ODE) of $x' = x^2 - t^2$ 
$\DeclareMathOperator{\dom}{dom}$Let $\gamma(t)$ be a the maximal solution of the diferential equation $x' = x^2 - t^2$ with initial condition $\gamma(0) = 0$. Show that $\gamma(t) \leq |t|$, for any $t \in \dom(\gamma)$ and conclude that $\dom(\gamma) = \mathbb{R}$.

Attempt: Assuming that $\gamma(t) \leq |t|$, for any $t \in \dom(\gamma)$, I was able to conclude that $\dom(\gamma) = \mathbb{R}$, using the fact that $\gamma (t)$ is a maximal solution.
Also, note that $(x^2 - t^2)' = 2x(x^2 - t^2) - 2t$. I tried to analyse for $t\geq 0$ and $t<0$, but there's not much I can infer about about $(x^2 - t^2)'$ since I don't know the sign of $x$. Is there any algebraic manipulation that could be helpful to procced with this idea?
Any help would be appreciated!
 A: $\def\d{\mathrm{d}}\DeclareMathOperator{\dom}{dom}$Suppose $x(t_0) > t_0$ for some $t_0 \in \dom(x) \cap (0, +∞)$ and define $t_1 = \inf\{t > 0 \mid x(t) > t\}$. Note that $x'(0) = (x(0))^2 = 0$, so the definition of $x'(0)$ implies that $|x(t)| < t$ in a neighborhood of $0$ and thus $t_1 > 0$. The definition of infimum implies that there exists a sequence $\{s_n\} \subseteq (t_1, +∞)$ with $s_n \searrow t_1$ as $n → ∞$ and $x(s_n) > s_n$ for all $n$. Thus the continuity of $x$ shows that $x(t_1) \geqslant t_1$. But if $x(t_1) > t_1$, then $x(t) > t$ for $t$ in a neighborhood of $t_1$, which is contradictory to the definition of $t_1$. So $x(t_1) = t_1$ and$$
x'(t_1) = \lim_{t → t_1} \frac{x(t) - x(t_1)}{t - t_1} = \lim_{n → ∞} \frac{x(s_n) - x(t_1)}{s_n - t_1} \geqslant 1,
$$
which is contradictory to the fact that $x'(t_1) = (x(t_1))^2 - t_1^2 = 0$. Therefore $x(t) \leqslant t$ for $t \in \dom(x) \cap (0, +∞)$. Analogously it can be proved that $x(t) \geqslant -t$ for $t \in \dom(x) \cap (0, +∞)$, and $|x(t)| \leqslant -t$ for $t \in \dom(x) \cap (-∞, 0)$.
