Let me write down the metric in the equatorial plane ($\vartheta = \pi/2$) of the Kerr space-time in Boyer-Lindquist coordinates:
$$
d s^2 = -\left(1 - \frac{2M r}{r^2 + a^2}\right) d t^2 + \frac{r^2+a^2}{r^2 - 2Mr + a^2} d r^2 + \left(r^2 + a^2 + \frac{2M r a^2 }{r^2 + a^2}\right) d \varphi^2 - \frac{2 M r a}{r^2 + a^2} dt d \varphi
$$
Now you have to trust me that going through the ring singularity at $r=0$ (remember that $r,\vartheta$ are in fact oblate ellipsoidal coordinates and $r=0$ is a disk) means going to $r$ negative. Now I would like to know if there is an $r$ such that the vector $\eta^\mu = \delta^\mu_\varphi $ is time-like. I find that
$$\eta^\mu \eta_\mu = g_{\varphi \varphi} = r^2 + a^2 + \frac{2M r a^2 }{r^2 + a^2}$$
which is negative in a certain range of negative $r$ (the range has a closed and cumbersome expression corresponding to two roots of a quartic equation). Since the integral curves of $\eta^\mu$ are closed, in that range of $r$, there are closed time-like curves. When you go off the equator and plot regions where $g_{\varphi\varphi}$ is negative, you find this region is a finite "doughnut" near the ring singularity in the $r<0$ part of the Kerr space-time.
The issue is, however, that if you start with your time-like curve in this doughnut, you can also leave it, go through the singularity back to $r>0$, and all the way to the inner horizon of the Kerr black hole. Then you return back to the $g_{\varphi\varphi}<0$ doughnut, and you are allowed to circle in $\varphi$ indefinitely with a small negative drift in $t$, and eventually end up closing your time-like curve (meet yourself from your proper past). So closed time-like curves are implied by the existence of the "acausal doughnut" all the way to the inner horizon of the Kerr black hole. This is also one of the reasons one often discards the region within the inner horizon as unphysical.