Is there a differential equation: $$\ddot x(t)=f(t, x(t)),\ f\in C^1(\mathbb{R}^2), \forall t\in \mathbb{R}\ f(t,0)=0;$$ that for any solution x(t) the following is true: $$\forall \epsilon>0\ \exists \delta>0\ (|x(0)|<\delta \implies \forall t\ge0\ |x(t)|<\epsilon)\ ? $$
I guess that there is no such an equation. I have the following intuition. Let's look at the equation thinking that f is a force and x(t) is a coordinate of a particle that moves under that force. If we provide sufficiently enough initial speed $$\dot x(0)$$ to the particle, which is not limited, the particle will reach any high coordinate no matter where it was at the initial moment of time, because force f does not depend on the speed of the particle.
I think you are right: Assume that there is such an equation. Then, for $\varepsilon =1$ there is some $\delta >0$ such that for any solution $|x(0)|< \delta$ implies $|x(t)|<1$ $(t \ge 0)$. Now set $$ c:= \max \{|f(t,\xi)|\mid (t,\xi) \in [0,1] \times [-1,1]\}. $$ Let $x:[0,\infty) \to \mathbb{R}$ be a solution with $|x(0)|< \delta$ and $x'(0)=1+ \delta + c/2$. Then, $|f(t,x(t))| \le c$ for each $t \in [0,1]$. Hence, for $t =1$ we have $$ 1 > |x(1)|= \left|x(0) +x'(0) + \int_0^1 \int_0^s f(\tau,x(\tau)) d\tau ds\right| $$ $$ \ge |x'(0)| - |x(0)| - \int_0^1 \int_0^s |f(\tau,x(\tau))| d\tau ds $$ $$ > |x'(0)| - \delta - \frac{c}{2} =1, $$ a contradiction.