= Solution
Work in the <Boyer-Lindquist coordinates> exterior $r>r_+=M+\sqrt{M^2-a^2}$, away from the axes. There $\Delta>0$, $\Sigma>0$, and
$$
g_{\phi\phi}=\frac{\sin^2\theta}{\Sigma}\left[(r^2+a^2)\Sigma+2Mra^2\sin^2\theta\right]>0.
$$
Inverting the $t,\phi$ block of the <Kerr metric> gives
$$
g^{tt}=-\frac{g_{\phi\phi}}D<0,\qquad D=g_{t\phi}^2-g_{tt}g_{\phi\phi}.
$$
Thus $W^a=-g^{ab}(dt)_b=-\nabla^at$ has $g(W,W)=g^{tt}<0$, and the smooth nonvanishing <timelike vector field> $W$ defines a <time orientation>. Choose it future-directed, matching increasing $t$ near infinity. For every nonzero future <causal vector> $Z$, the <Lorentzian inner product> satisfies $g(W,Z)<0$, hence
$$
\boxed{dt(Z)=-g(W,Z)>0.}
$$
So $t$ increases strictly along every regular future <causal curve> in this exterior. Notice that $\partial_t$ itself can be spacelike in the <Kerr ergoregion>; the timelike object used here is $-\nabla t$. Direct substitution also gives $D=\Delta\sin^2\theta$.
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