Completeness and Topological Equivalence How can I show that if a metric is complete in every other metric topologically
equivalent to it , then the given metric is compact ?
Any help will be appreciated .

I encountered this result in Queffélec's book's Topologie. The proof is due to A.Ancona.
It's known as Bing's theorem. We can assume WLOG that $d\leq 1$, otherwise, replace $d$ by $\frac d{1+d}$. We assume that $(X,d)$ is not compact; then we can find a sequence $\{x_n\}$ without accumulation points. We define 
$$d'(x,y):=\sup_{f\in B}|f(x)-f(y)|,$$
where $B=\bigcup_{n\geq 1}B_n$ and 
$$B_n:=\{f\colon X\to \Bbb R,|f(x)-f(y)|\leq \frac 1nd(x,y)\mbox{ and }f(x_j)=0,j>n\}.$$
Since $d'\leq d$, we have to show that $Id\colon (X,d')\to (X,d)$ is continuous. We fix $a\in X$, and by assumption on $\{x_k\}$ for all $\varepsilon>0$ we can find $n_0$ such that $d(x_k,a)>\varepsilon$ whenever $k\geq k_0$. We define 
$$f(x):=\max\left(\frac{\varepsilon -d(x,a)}{n_0},0\right).$$
By the inequality $|\max(0,s)-\max(0,t)|\leq |s-t|$, we get that $f\in B_{n_0}$. This gives equivalence between the two metrics. 
Now we check that $\{x_n\}$ still is Cauchy. Fix $\varepsilon>0$, $N\geq\frac 1{\varepsilon}$ and $p,q\geq N$. Let $f\in B$, and $n$ such that $f\in B_n$. 


*

*If $n\geq N$, then $|f(x_p)-f(x_q)|\leq \frac 1nd(x_p,x_q)\leq \frac 1n\leq \varepsilon$;

*if $n<N$ then $|f(x_p)-f(x_q)|=0$.

