Past exam of the options II course of the University of Oxford / 2009 / E / E3 / b / iv

Every point of $A$ differs from $a$, so its singleton is an open ball by part (ii). The union of these singleton open sets is $A$. Thus $A$ is open whenever $a\notin A$.

If $A$ is open and contains $a$, it contains some $d_a$-ball about $a$, which is exactly the corresponding $d$-ball. Conversely, if $B_d(a,\epsilon )\subseteq A$, this is a $d_a$-neighbourhood of $a$, while every other point of $A$ has its open singleton inside $A$. Hence $A$ is open. Thus the stated ball-containment condition is necessary and sufficient.