let $(x(t), y(t))$ be a solution to the system $x'=y-x^3, y'=-x-y^3$.
I want to prove that $(x(t), y(t))$ is bounded for $t>0$
What I did:
from the equations comes $0\ge -(x^4+y^4)=x'x+y'y=\frac d{dt}(\frac{x^2+y^2}2)$, so $\frac{x^2+y^2}2$ decreases. So $\sqrt{x^2+y^2}$ also decreases. this means that $||(x,y)||$ decreases and so $||(x(t),y(t))||\leq ||(x(0),y(0))||$ for all $t>0$.
the above proves what I want, but why should $(x(t), y(t))$ be defined at $t=0$?
why should $(x(t),y(t))$ be defined at $t=0$?
Because this is (or should have been) a starting assumption. Without it, the statement is false. Starting from an arbitrary point $(x_0,y_0)\ne (0,0)$ and moving backward in time, you'll see the solution blowing up at finite time, because the derivative grows at least quadratically: $$-\frac{d}{dt}(x^2+y^2) = 2(x^4+y^4) \ge (x^2+y^2)^2$$ Since the system is autonomous, with a time shift you can place the blow-up time at $t=0$. This creates a solution on $(0,\infty)$ that is unbounded near $0$.
