A complete Riemannian manifold admits cutoff functions with uniformly bounded first derivatives 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.
 A: 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.
A: This is not an answer but a long comment. Take a look at the proofs of Theorem 1 and Corollary 3 in this paper as they construct smooth Lipschitz approximation of Lipschitz functions on Riemannian manifolds, the approximation scheme they use might work in your setting as well. The result you need might be also contained in the earlier papers by Greene and Wu (see the references 7-9 in the link). 
