I'm reading a paper which uses the following fact; it appears to be standard but I am not sure where to look for a proof.
Claim. Let $M$ be a complete Riemannian manifold (assumed to be second countable, so no long lines). There is an increasing sequence of open sets $U_n$ with $\bigcup_n U_n = M$ and smooth, compactly supported functions $\phi_n : M \to [0,1]$ such that $\phi_n = 1$ on $U_n$ and $\sup_n |\nabla \phi_n| < \infty$.
This is trivial if $M$ is compact (take $U_n = M$ and $\phi_n = 1$). If we drop the requirement that the derivatives of $\phi_n$ be uniformly bounded, it's a consequence of the $\sigma$-compactness of $M$ and Urysohn's lemma. Also, the completeness is essential as we can see by taking $M$ to be an open interval.
I would appreciate a proof (or at least a hint) or a reference.
One idea is to filtrate $M$ with distance balls. Here's a sketch.
Suppose $M$ is (connected) noncompact and complete. Let $p\in M$. Put $U_n = B(p,n)$. Note $U_n \subset U_{n+1}$.
Completeness implies the containment is proper: [1] By noncompactness, $U_i$ is compactly contained in $M$ (otherwise $M$ would be closed and bounded). Let $x_i\in \partial U_i$ ($x_i$ should also be in the closure of the interior of the complement of $U_i$). Connect $p$ to $x_i$ by a geodesic and by completeness we may extend the geodesic slightly. The extension is contained in $U_{i+1}$ but not in $U_i$.
Now since you're guaranteed that the distance between $\partial U_n$ and $\partial U_{n+1}$ is always equal to $1$, you can control the magnitude of the derivative of $\phi_n$.
For those reading along, the difficulty seems to be in explicitly constructing the $\phi_i$ as cutoff functions.
[1] Here's an alternative construction of the $U_i$. By noncompactness choose a sequence $x_i$ with no convergent subsequence. By completeness (metric completeness is equivalent to geodesic completeness for Riemannian manifolds), for any $p\in M$ fixed, $d(p,x_i)$ increases without bound. Remove points from the sequences so that $d(p,x_i)$ is monotone increasing and for no $i,j$ is $d(x_i,x_j) < 1$. Choose $U_i = B(p,d(p,x_i))$. Again since the distance between the $U_i$ and $U_{i+1}$ is no less than $1$, you get uniform control over the gradient of the cutoff function.