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This question already has an answer here:

Which metrizable topological spaces $(X,\tau)$ posses the following property:

Every compatible metric (i.e one which induces the same topology $\tau$) is complete.

Compact metrizable spaces satisfy this.

Are there any non-compact examples?

Note:

It turns out that if you impose additional structure the answer is yes. In particular, if $M$ is a smooth manifold, and every Riemannian metric on $M$ is complete, then $M$ must be compact.

Reference: Nomizu, Katsumi, and Hideki Ozeki. "The existence of complete Riemannian metrics." Proceedings of the American Mathematical Society 12.6 (1961): 889-891.

http://www.jstor.org/discover/10.2307/2034383?uid=2&uid=4&sid=21105114015163is

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marked as duplicate by user642796 Aug 1 '15 at 20:05

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    $\begingroup$ I'd guess that the answer is "only compact spaces", since otherwise you can take some sequence without a convergent subsequence, and force the metric to make it into a Cauchy sequence somehow. $\endgroup$ – Asaf Karagila Aug 1 '15 at 19:59