Proving that every countable metric space is disconnected? Any hints on how I should go about this? Thanks.
 A: Do you mean totally disconnected? In that case, take two points $x, y$ with $d(x, y) = a$. Then, since there are only countably many points in the space, and there are uncountably many real numbers less than $a$, there must be a distance $b < a$ such that there are no points exactly at a distance $b$ from $x$.
That means that the ball around $x$ with radius $b$ is both open and closed, and so is its complement. Thus you have partitioned the space into two disjoint clopen sets, one containing each point. Since the points were arbitrary, this means that the space is totally disconnected.
A: The statement is false if $|X|=1$, so countable here is probably supposed to be understood as countably infinite. 
HINT: Let $X$ be the space, and let $x,y\in X$ with $x\ne y$. Let $r=d(x,y)>0$. Use the fact that $X$ is countable to show that there must be an $s\in(0,r)$ such that 
$$\{z\in X:d(x,z)=s\}=\varnothing\;.$$
If $B(x,s)=\{z\in X:d(x,z)<s\}$, show that $B(x,s)$ and $X\setminus B(x,s)$ are non-empty open sets in $X$, and conclude that $X$ is not connected.
