Cofinite topology on infinite sets can't be induced from a metric space. I'm a very beginner in topology, and I have a question. 
The book I'm studying says that cofinite topologies on any infinite sets can't be induced from any metric spaces, but how do I show this? 
I haven't learned anything about hausdorf, compactness, separability, or connectedness yet. 
I feel that I need to use the fact that I can't make any open ball in (X,d) whose complement is finite. In here X is a set and d is a metric given.
 A: As Randall says, Hausdorffness is the way to go, but there's no need to say you're using Hausdorffness. You can just give the following Hausdorffness inspired proof.
Assume for contradiction that there is an infinite set $X$ with metric $d$ such that the induced topology on $X$ is the cofinite topology. Let $x,y$ be distinct points of $X$. Then let $\varepsilon=d(x,y)>0$. Let $U=B_{\varepsilon/2}(x)$, $V=B_{\varepsilon/2}(y)$. By the triangle inequality, $U\cap V=\varnothing$. However, $U$ and $V$ are nonempty open sets, hence $U$ and $V$ are both cofinite. Hence $(U\cap V)^C = U^C\cup V^C$ is also finite, so $U\cap V$ is cofinite. In particular, it is nonempty. Contradiction.
A: In the cofinite topology on the infinite set $X$, any two non-empty open sets have a non-empty intersection. This should be reasonably clear:  if $U$ and $V$ are non-empty and open and $U \cap V$ is empty, then 
$$
X = X -(U \cap V) = (X-U) \cup (X-V).
$$
But now the infinite set $X$ is a union of two finite sets, a contradiction.
Now, in a metric space, do ALL pairs of non-empty open sets always have non-empty intersection?
