Proving $|x(t)|\lt |t|$ for the solutions $x$ of an ODE The problem is:
Let $x(t)$ a solution of $$x'=-x(t^2-x^2),$$ so that $|x(t_0)|\lt |t_0|$. Show that for all $|t|\gt |t_0|$, $|x(t)|\lt |t|$.
Suppose that $t_0\geq 0$. Consider

Then $(t_0,x(t_0))$ is over the red line. I don't see why: $x$ solution of the above ODE implies that the graph of $x$ stay between the blue lines to the right of the red line and to the left of the orange line. 
Attempt.
The attempt becomes an answer...
 A: For simplicity, multiply the equation by $2x$ and subtract $2t$ to get
$$
(x^2-t^2)'=2x^2(x^2-t^2)-2t\tag{1}
$$
First, let's deal with $t>0$.
$(1)$ says that if $x^2-t^2<0$, then we also have $(x^2-t^2)'<0$. Therefore, if $x(t_0)^2-t_0^2<0$, we have $x(t)^2-t^2<0$ for all $t\ge t_0$.
Next, let's deal with $t<0$.
$(1)$ says that if $x^2-t^2\ge0$, then we also have $(x^2-t^2)'\ge0$. Therefore, if $x(t)^2-t^2\ge0$, we have $x(t_0)^2-t_0^2\ge0$ for all $t_0\ge t$. The contrapositive says that for all $t\le t_0$, if $x(t_0)^2-t_0^2<0$, then we have $x(t)^2-t^2<0$.
A Graphical Explanation
Consider the direction field (aka slope field and flow diagram) for the solutions. That is, the vector field $\color{#0000c0}{\frac{\mathrm{d}}{\mathrm{d}t}(t,x)}$ which is tangent to all solutions $\color{#c00000}{(t,x)}$.
$\hspace{3cm}$
The important part to consider is the flow across the boundary $x^2=t^2$ between the regions where $\color{#00c000}{x^2< t^2}$ and $\color{#c000c0}{x^2>t^2}$. Notice that the on the boundary, the flow is horizontal since $x'=x(x^2-t^2)=0$.
$\hspace{4.2cm}$
Thus, when $t<0$ (on the left), a solution can only move from the region where $\color{#00c000}{x^2< t^2}$ to the region where $\color{#c000c0}{x^2>t^2}$. When $t>0$ (on the right), a solution can only move from the region where $\color{#c000c0}{x^2>t^2}$ to the region where $\color{#00c000}{x^2< t^2}$.
A: Consider first $t_0\geq 0$.
Suppose that the set $$C:=\{t\gt t_0:x(t)\geq t\}$$ is not empty, thus, since $C$ is bounded below by $t_0$, $$\tau=\inf C$$ is well defined. We can proof (by taking a sequence $C\ni t_n\to\tau$ and assuming continuity of $x$) that $\tau\in C$. In particular $\tau\gt t_0$ and then $$x(\tau)\geq\tau\gt 0,$$ then, there exist an $\alpha\gt 0$ such that $x$ is positive over $]\tau-\alpha,\tau+\alpha[\subset]t_0,\infty[$.
Then for all $t\in]\tau-\alpha,\tau[$, by the definition of $\tau$,  $t\notin C$ (otherwise $t\in C$ with $t\lt\inf C$), so $0\lt x(t)\lt t$ and then $$x'(t)=x(t)(x(t)^2-t^2)\lt 0,$$
so $x$ is decreasing over $]\tau-\alpha,\tau[$. 
Pick $t\in ]\tau-\alpha,\tau[$, then $$x(t)\geq x(\tau)\geq\tau\gt t,$$ thus $t\in C$, and that's contradictory. 
Therefore $C=\emptyset$ and then for all $t\gt t_0$, $x(t)\lt t$.
A: Hint: If $x(t_0) &lt |t_0|$ then, by the differential equation,
$x'(t_0) = -x(t_0)(\textrm{something positive})$. 
A: If $x(t_0)=0$ then $x(t)=0$ for all $t\ge t_0$. Without loss of geneality, we assume $x(t_0)\ne0$.
If $0&ltx(t_0)&ltt_0$, you want to show that $0&ltx(t)\le t$ for all $t\ge t_0$. Now, $x'(t_0)=-x(t_0)(t_0^2-x(t_0)^2)&lt0$. By uniqueness of solutions $x(t)>0$ for all $t\>t_0$. And since the right hand side of the equation is negative on the region $\{(t,x):0&ltx&ltt,t>t_0\}$, it is easy to see that $x$ is decreasing.
If $x(t_0)=t_0$ then $x'(t_0)=0$. Taking derivatives in both sides of the equation we get $x''(t_0)=-2\,t_0\,x(t_0)&lt0$. Thus $x'(t)&lt0$ on some interval $(t_0,t_0+\delta)$, $\delta>0$, $x(t)&ltt$ on that same interval, and the argument in the previos paragraph carries through.
I leave to you the case $-t_0\le x(t_0)&lt0$.
