Let $M$ be a smooth manifold and $X\in\mathfrak{X}(M)$. Suppose there is a positive, proper function $f\in C^\infty(M)$ such that $df(X)$ is bounded. Prove that $X$ is complete.
Take $\gamma: I\to M$ a maximal integral curve of $X$. Then $(f\circ\gamma)'(t)=df_{\gamma(t)}(\gamma'(t))=df_{\gamma (t)}(X_{\gamma(t)})$ is bounded.
My idea is to prove $f\circ\gamma(I)$ is contained in a compact, so that $\gamma(I)$ is also contained in a compact (since $f$ is proper). By the escape lemma (see Lee's Introduction to Smooth Manifolds, lemma $12.11$), $I=\mathbb{R}$, as desired.
The problem is that I can't see how to prove this from the fact that $(f\circ \gamma)'$ is bounded.
Any ideas?
$$\begin{align*} (f\circ \gamma) (t) &= (f\circ \gamma)(0) + \int_0^t (f\circ \gamma)'(s) ds. \end{align*}$$
So
$$|(f\circ \gamma)(t)| \le |(f\circ \gamma)(0)| + C|t|,$$ where $C$ is the bound of $df (X)$.
If $I$ is bounded, then so is $(f\circ \gamma)(I)$. Thus $\gamma(I)$ is also bounded since $f$ is proper.
Edit: If $I$ is of the form $(-\infty, a)$, one can then restrict to $(c, a)$ and using the boundedness to claim that it can be extended to $(c, b)$ for all $b>a$. Similar for $(a, \infty)$.