Let $X=\mathbb{R}^*$ and $\forall(x,y)\in X^2$, $d(x,y)=|x-y|+|x^{-1}-y^{-1}|$. Is $(X,d)$ a complete metric space ?
Let $(a_n)$ be a Cauchy sequence on $(X,d)$. Then $\forall\epsilon>0,\exists n_\epsilon\in\mathbb N,p,q\geq n_\epsilon\implies d(a_p,a_q)<\epsilon$. This implies that $(a_n)$ is also Cauchy on $(X,d_1)$ and $(X,d_2)$ where $d_1$ is the usual metric on $\mathbb R$ and $\forall(x,y)\in X^2, d_2(x,y)=|x^{-1}-y^{-1}|$.
From there, I'm still debating if I want to find a counter example showing the initial space is not complete or show that is a sequence is Cauchy in the original space with the added condition that it cannot possibly converge to $0$ in which case $(X,d)$ should be complete. I'm not sure if it can help, but I was previously asked if the graph of $x\in X\mapsto x^{-1}$ was closed in $(\mathbb R^2,\delta)$ where $\delta$ is the usual metric on $\mathbb R^2$.
Hint: $(a_n)$ and $(\frac 1 {a_n})$ are Cauchy in the usual metric and this implies they are both bounded. The first sequence converges to some real number $a$ and $a \neq 0$ because the second sequence is bounded. Can you complete?