Finite Metric Space 
In finite metric space all sets are open 

Is it true because all singletons are open and the union of them is open too?
 A: It's true, but another way:
In any metric space, finite or not, all singletons are closed. So finite sets are closed.
In a finite metric space, any subset has a finite (hence closed) complement. So all sets are open.
A: Let $M=\{x_1,x_2,\cdots,x_n\}$ be a finite metric space. 
Let $A=\{x_1\}$ Then $A^c=\{x_2,x_3,\cdots,x_n\}$ is a finite set, hence closed. Therefore, $A$ is open.
Similarly, all singletons are open. 
Now, let $B$ be any subset of $M.$ Then $B$ is a finite union of open sets(singletons), hence open.
In general, all subsets of a metric space are open if singletons are open.
Edit 1: To show a finite set is closed, it is sufficient to show that any singleton is closed (since finite union of closed sets is closed)
Let $X$ be a metric space and $x \in X$. We need to show $\{x\}$ is closed. It suffices to show that $\{x\}^c$ is open. 
Let $ y \in \{x\}^c.$ Then $y\neq x.$ Hence, $d(x,y)>0.$ Let $$r=\frac{d(x,y)}{2}>0$$
Then $B(y,r) \subseteq \{x\}^c.$ This shows $\{x\}^c$ is open.
Edit 2: Let $z \in B(y,r)$. Then $$d(z,y)<r=\frac{d(x,y)}{2}<d(x,y)$$
Hence, $z \neq x$. Therefore, $z \in \{x\}^c.$
As $z \in B(y,r)$ was arbitrary, therefore $B(y,r) \subseteq \{x\}^c.$
