 Solution

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.