show that $(X,d )\ $ is compact . if every proper closed subset of a metric space $(X,\ d)\ $  is finite.
show that $(X,d )\ $ is compact .
how to show ?? by definition or by some equivalent criteria. Any hint 
 A: Let $(A_\lambda)_{\lambda\in\Lambda}$ be an open cover of $X$. Fix a $\lambda_0\in\Lambda$ such that $A_{\lambda_0}\neq\emptyset$. Then $X\setminus A_{\lambda_0}$ is a proper closed subset of $X$ and therefore it is finite. So $X\setminus A_{\lambda_0}=\{x_1,x_2,\ldots,x_n\}$. For each $k\in\{1,2,\ldots,n\}$, let $\lambda_k\in\Lambda$ be such that $x_k\in A_{\lambda_k}$. Then$$\{x_1,x_2,\ldots,x_n\}\subset\bigcup_{k=1}^nA_{\lambda_k}$$and therefore$$X=\bigcup_{k=0}^nA_{\lambda_k}.$$
A: Extra credit:  Show that $X$ is actually finite.  
Proof:  If $X$ is empty or has only one point, fine.  Otherwise choose two points $x$ and $y$.  Since $X$ is a metric space, it is Hausdorff, so the two points have disjoint open neighborhoods. Conclude those open neighborhoods are finite.  Hence $x$ is an isolated point. We have now shown every point is isolated.  Again, choose a point $x$.  Then $\{x\}$ is an open set, hence its complement is finite.  
We only used that the space is Hausdorff.  So:

Proposition:  Let $X$ be a Hausdorff topological space in which every proper closed set is finite.  Then $X$ is finite.

