The question in the title seems a natural one to ask, and I suspect that it is already considered, even completely solved, somewhere in the literature. Although I prefer some explanation of the idea(s), relevant references are also appreciated.
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4$\begingroup$ This seems to be answered here: math.stackexchange.com/questions/1904941/…. $\endgroup$– Leo MoosCommented Feb 2, 2023 at 10:04
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This is not hard to prove. The idea of a proof is as follows. Take any metric $g$ on a manifold $M$. Define $d:M\to \mathbb{R}_+$ by saying that $d(x)$ is the infimum of all lengths of curves $\gamma:[0,b)\to M$ with $\gamma(0)=x$, such that $\gamma$ is proper (In particular: For every compact subset K\subset M, there is a $t\in [0,b)$ with $\gamma(t)\not\in K$). Choose a smooth function $f:M\to (0,1]$ with $0<f<d$. Then $\bar g:= f^{-2}g$ is complete.
There is much more possible, namely bounded geometry. This was published in
Müller, Olaf; Nardmann, Marc;
Every conformal class contains a metric of bounded geometry.
Math. Ann. 363 (2015), no. 1-2, 143–174.
