$a:[0,\infty)\to \mathbb{R}$ is a continous and bounded and
$$x'(t)\ =\left(\begin{matrix}0&1\\-a(t)&0\end{matrix}\right) \ x(t)$$ has a non-zero solution like $y(t)$ such that $\lim_{t \to \infty} y(t)=0$.
Show that this equation has an unbounded solution on $[0,\infty)$.
Let $X=(x_1,x_2)$ and $y=(y_1,y_2)$. The system can be written as he second order linear equation $$ x_1''+a(t)\,x_1=0. $$ We know that it has a solution $y_1$ with $\lim_{t\to\infty}y_1(t)=\lim_{t\to\infty}y_1'(t)=0$. To find another solution let $x_1=z\,y_1$. Then $$ z''\,y_1+2\,z'\,y_1'+z\,y_1''-a\,z\,y_1=0\implies \frac{z''}{z'}=-2\,\frac{y_1'}{y_1}\implies z'=\frac{C}{y_1^2}.\tag{1} $$ Thus $$ x_1(t)=y_1(t)\int_0^t\frac{ds}{(y_1(s))^2} $$ is a solution of equation (1) (and the firs component of ta solution of the system.) I leave it to you to show that $\lim_{t\to\infty}x_1(t)=\infty$.