Question about complete metric on manifolds

Question

I've recently been wondering about whether non-complete metrics on manifolds can be transformed into complete metrics on manifolds and whether all manifolds have complete metrics. After some googling I came across this link and the first comment says that any metric is actually conformal to a complete metric. I was wondering if anybody can show me a proof of this because I have had difficulty finding one. Thank you!

Answer 1

$\textbf{Theorem 1}$ of The Existence of complete Riemannian Metrics is what you're looking for :

For any Riemannian metric $g$ on $M$ there exists a complete Riemannian metric which is conformal to $g$

The proof starts right on the first page.