The analysis textbook we use in class states
If $(X,d)$ is a metric space $\{a_0\}$, $a_0\in X$, is not open, if $X\neq\{a_0\}.$
This is clearly wrong taking the discrete metric as a counterexample. I am wondering, is there any other metric on any set (which is not equivalent to the discrete metric) which has at least one open singleton? Or are all singletons in a metric space necessarily open if one singleton is open and hence the metric equivalent to the discrete metric?
Edit: José has given a nice example but I am still wondering what happens if one puts the additonal constraint on $X$ of being connected. Does in this case $d$ have to be equivalent to the discrete metric?
Take $[0,1]\cup\{2\}$ endowed with the usual metric. This space has one and only one open singleton, which is $\{2\}$.