= Solution
Along the downward <gradient flow>, the <gradient-flow dissipation identity>, the <chain rule> and the defining property of the <Riemannian gradient> give
$$
\frac{d}{dt}f(\gamma_x(t))=-\|\nabla f(\gamma_x(t))\|^2.
$$
A smooth <vector field> on a <compact manifold> is complete, so the trajectory exists for every $t\geq0$. In particular, its function value never increases.
If the trajectory is still in $C_{[a,b]}$ at time $t$, its monotone function value has stayed in $[a,b]$ throughout $[0,t]$. Consequently
$$
a\leq f(\gamma_x(t))=f(x)-\int_0^t\|\nabla f(\gamma_x(s))\|^2\,ds\leq b-\varepsilon^2t.
$$
Thus \b[a uniform exit bound] is
$$
\boxed{T=(b-a)/\varepsilon^2.}
$$
The trajectory is outside the closed band for every $t>T$. Once it leaves below level $a$, monotonicity prevents its return. The argument also covers the zero-width band $a=b$.
For the <total occupation-time bound for a gradient flow>, let $m=\min_M f$ and $F=\max_M f$, which exist by <compactness>. For every $R>0$, the same <gradient flow> identity gives
$$
\varepsilon^2\int_0^R\mathbf1_{C_\varepsilon}(\gamma_x(t))\,dt
\leq\int_0^R\|\nabla f(\gamma_x(t))\|^2\,dt
=f(x)-f(\gamma_x(R))\leq F-m.
$$
Taking $R\to\infty$, by <monotone convergence theorem>, gives the \b[total occupation-time bound]
$$
\boxed{\int_0^\infty\mathbf1_{C_\varepsilon}(\gamma_x(t))\,dt\leq\frac{\max_M f-\min_M f}{\varepsilon^2}.}
$$
This allows arbitrarily many visits to $C_\varepsilon$; it bounds their total duration. Neither of these two estimates needs the <Morse function> hypothesis.
For <exponential convergence of a Morse gradient flow>, first observe that a limiting point $z=x_\infty$ is a <critical point>. Indeed, writing $\phi_s$ for the flow and using continuous dependence on initial data,
$$
\phi_s(z)=\lim_{t\to\infty}\phi_s(\gamma_x(t))=\lim_{t\to\infty}\gamma_x(t+s)=z.
$$
Differentiation at $s=0$ gives $\nabla f(z)=0$.
We use the <Morse lemma>: near a <nondegenerate critical point> of <Morse index> $\lambda$, there are coordinates $(u,v)\in\mathbb R^\lambda\times\mathbb R^{n-\lambda}$ in which
$$
f(u,v)=f(z)-|u|^2+|v|^2.
$$
Choose a <Riemannian metric> equal to the Euclidean metric on a smaller ball in these coordinates. Such a global <Riemannian metric> exists by a <partition of unity>: patch this metric to any background metric outside a slightly larger ball. One can do this simultaneously at every <critical point>, since a <Morse function> on a <compact manifold> has only finitely many <critical points>. The metric is chosen before forming its <gradient flow>; changing the metric can change trajectories.
For this metric, any trajectory converging to $z$ eventually remains in the smaller coordinate ball. From some time $t_0$ onward its equations are exactly
$$
\dot u=2u,\qquad\dot v=-2v.
$$
Convergence forces $u(t_0)=0$, and then
$$
u(t)=0,\qquad v(t)=e^{-2(t-t_0)}v(t_0),\qquad t\geq t_0.
$$
The straight radial segment stays in the coordinate ball, so its length bounds the <Riemannian distance> from above:
$$
d(\gamma_x(t),z)\leq |v(t_0)|e^{-2(t-t_0)}.
$$
Increase $C$ beyond both $e^{2t_0}|v(t_0)|$ and $\max_{0\leq t\leq t_0}e^{2t}d(\gamma_x(t),z)$. Then the requested strict estimate holds for all $t\geq0$:
$$
\boxed{d(\gamma_x(t),x_\infty)<Ce^{-2t}.}
$$
This proof uses the <Morse lemma>, existence and uniqueness for smooth <ordinary differential equations>, continuous dependence on initial data, and the existence of a <partition of unity>. The expanding coordinates explain why convergence to a saddle occurs only along its <stable manifold>.
For a counterexample, use an arc coordinate $x$ around a point of the circle, its flat local metric, and a globally smooth function that equals $x^4$ on this arc. Such a function can be made by a smooth cutoff, agreeing with a positive constant outside a larger arc. For $x_0>0$ sufficiently small the downward <gradient flow> remains in the arc and is
$$
\dot x=-4x^3,\qquad\boxed{x(t)=\frac{x_0}{\sqrt{1+8x_0^2t}}.}
$$
It converges to the degenerate <critical point> $0$, and its <Riemannian distance> from $0$ is $x(t)$ for sufficiently large $t$. For every $k>0$, $e^{kt}x(t)\to\infty$, so no exponential bound is possible. This counterexample is on a <compact manifold> as well.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-116-gradient-decay.png]
{title=Exponential decay near a Morse minimum and algebraic decay near a degenerate quartic minimum, on a logarithmic vertical scale}
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