Can a nonautonomous vector field on $\mathbb{R}$ that has no equilibrium points have periodic orbits?
His answer:
"Yes. Consider the example $\dot x = \cos (t)$. A solution is $x(t) = \sin t$, which is periodic in time."
Surely $\dot x = \cos (t)$ has equilibrium points at $t=\pi/2 + k\pi, k\in\mathbb{Z}$ ?
An equilibrium point is some $x_0\in\mathbb{R}$ such that $x(t) =x_0$ is a solution for all $t$. This is not what you have provided, and so it is not an equilibrium point. Moreover, you should realise that must solve a differential equation over a continuous interval, not at specific points in the domain.