$X$ be a set such that any two metric on $X$ are equivalent ; then is it true that $X$ has to be countable? Can $X$ be countably infinite? Let $X$ be a set such that any two metric on $X$ are equivalent (i.e., generates the same topology i.e. any metric on the set generates discrete topology); then is it true that $X$ has to be countable? Is there an example of a countably infinite such set $X$ such that any two metrics on $X$ are equivalent? 
 A: On any $X$ we can define the discrete metric. As soon as $X$ is infinite, we can define a metric that induces a topology that is not discrete: Pick a countably infinite subset $$N=\{x_0,x_1,x_2\ldots\}\subseteq X$$ and define a metric that makes $N$ "look like" $\{0\}\cup\{1,\frac12,\frac13,\frac14,\ldots\}$ and everything alse at distance $1$.
More precisely, with $f\colon \Bbb N_0\to \Bbb R$ given by $$f(n)=\begin{cases}\frac1n&\text{if $n>0$}\\0&\text{if $n=0$}\end{cases}$$
let
$$d(x,y)=\begin{cases}\left|f(n)-f(m)\right|&\text{if $x,y\in N$ with $x=x_n,y=x_m$}\\0&\text{if $x=y\notin N$}\\1&\text{if $x\ne y$ and ($x\notin N$ or $y\notin N$)}\end{cases} $$
Verify that this is indeed a metric and that under this metric the set $\{x_0\}$ is not open.
A: This answer is a side story. In fact, your $X$ should be finite. You can easily check that every metric on finite set generates the discrete topology. (In general, every Hausdorff topology on finite set is discrete.) It is a consequence of von Eitzen's answer. 
However implying finiteness of $X$ from von Eitzen's requires the axiom of choice and choice is necessary:
Let $X$ be an amorphous set and $x$ be a metric on $X$. 
Fix $x_0\in X$ and consider the image of $X$ under the function $$x\mapsto d(x,x_0).$$
 It is a function from an amorphous set to a linearly ordered set so the image must be finite. Therefore there is a minimal distance $r$ from $x_0$ to another points in $X$. It says that the ball centered at $x_0$ with radius $r/2$ is the singleton $\{x_0\}$. We can take $x_0$ arbitrarily so $(X,d)$ is discrete.
