Why is indiscrete topology unmetrizable? 
For instance, the indiscrete topology for $X$ cannot arise from a metric when $X$ has more than one point. One way to see this is to note that the complement of a one-point set in a metric space is always open.

I don't see what is going on here. Say $X=\{a,b\}$. Then the indiscrete topology for $X$ would be $\mathfrak{F}=\{\emptyset,X \}$. But now what?
 A: Let $\langle X,d\rangle$ be a metric space, and let $x\in X$. If $y$ is any other point of $X$, then $d(x,y)>0$. Let $r_y=d(x,y)$; then $B_d(y,r_y)$ is an open ball about $y$ that does not contain $x$. Now let
$$U=\bigcup_{y\in X\setminus\{x\}}B_d(y,r_y)\;.$$


*

*$U$ is a union of open balls, so $U$ is open.  

*If $y\in X\setminus\{x\}$, then $y\in B_d(y,r_y)\subseteq U$, so $U\supseteq X\setminus\{x\}$.  

*For all $y\in X\setminus\{x\}$, $x\notin B_d(y,r_y)$, so $x\notin U$.


The last two points show that $U=X\setminus\{x\}$; couple that with the first, and you see that $X\setminus\{x\}$ is open. In other words, we’ve just proved the claim that you quoted: 

the complement of a one-point set in a metric space is open.

Now apply that to your space $X=\{a,b\}$ with the indiscrete topology $\{\varnothing,X\}$: $\{a\}$ is a one-point subset of $X$, and its complement is $\{b\}$, which is not open. Therefore the topology on $X$ cannot come from any metric. And the same argument works for the indiscrete topology on any set with more than one point.
A: What you need to do now is assume there exists a metric $d$, and derive a contradiction.  For instance, if $X = \{a,b\}$, then there is some distance $D = d(a,b) > 0$.  What can you now say about the (open) ball of radius $D$ centered at $a$?  In other words, which points $x\in X$ satisfy $d(a,x) < D = d(a,b)$?
