Where can I find proof of the following classical fact?

The existence of a nontrivial Killing vector field on a compact Riemannian manifold $M$ is equivalent to the existence of a nontrivial $\Bbb S^1$-action on $M$.

Is there any counterexample in non-compact case?

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    $\begingroup$ For the second question, consider the manifold $\mathbb{R}$ with the standard metric has a killing vector field $\frac{\partial}{\partial x}$, but $\mathbb{R}$ has no nontrivial continuous $S^{1}$-actions. To see this, note that any non-trivial orbit of an $S^{1}$-action is homeomorphic to $S^{1}$, but no subset of $\mathbb{R}$ is homeomorphic to $S^{1}$. The first question is very interesting and I hope it gets answered! $\endgroup$ – Nick L Dec 14 '18 at 15:38

The group $\text{Isom}(M)$ has the form of a Lie group; if $M$ is compact, this Lie group is compact.

Given any non-trivial Killing vector field, this gives a map $X: \Bbb R \to \text{Isom}(M)$; its image is a commutative subgroup. The closure of the image of $X$ is still commutative (the equation $ab = ba$ is true on an open dense subset of $\bar X \times \bar X$) and still a subgroup. If $M$ is compact, this subgroup must then be compact; because $\bar X$ is a compact, connected, non-trivial abelian Lie group we thus have $\bar X \cong T^n$ for some $n>0$. In particular, there is a circle subgroup of $\text{Isom}(M)$, and hence a faithful action of $S^1$ on $M$ by isometries.

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