Topology, open sets? If $(X,d)$ is a metric space, $X=\{1,2\}$ and a topology $T=\{ ∅,\{1\},\{2\},\{1,2\}\}$. Is it correct to say that each element of $T$ is an open set of $X$, for any $d$? What if $X$ had no metric at all?
 A: Once you include the empty set as suggested in the comments, this is a topology. In particular, this is the discrete topology. In general, metrics on finite sets are a little strange. One such metric that could be used to generate the discrete topology on any set (including finite ones) is the (obviously enough) discrete metric.
https://en.wikipedia.org/wiki/Discrete_space
You can prove actually that finite metric spaces are discrete. This is a good exercise. Try examining what happens for different size balls at different points.
It's worth nothing that the topology is somehow prior to the metric though. That is to say, topological structure doesn't require metric structure, but a particular topological structure might coincide or be induced with/by a metric.
A: It is true that if $(X, d)$ is metric space and $X$ is finite, then all sets in $X$ are open: 
For each $a\in X$. Let $r = \frac 12 \min_{x\in X} d(a, x) >0$. Then 
$$\{a\} = B(a, r)$$
is open in $X$. Actually all $T_1$ finite space has this property. 
If $X$ has no metric and has more than one element, it might not be (think of trivial topology). 
