I think that I was a little confused about the assumption that $X$ is $\sigma$-compact. Expressing $X$ as a disjoint union $X = \bigcup_{\alpha \in A} X_{\alpha}$, where $X_{\alpha}$ is clopen and separable, the new metric $d'$ defined by $d'(x,y) = d(x,y)$ if $x,y$ are in the same component and $d'(x,y) = 1$ if $x,y$ are in the differents components is equivalent to $d$ and $(X,d')$ is complete.

So, we can conclude that a locally compact metric space is separable... very cool. Thanks