 Solution

An isolated point outside $B$ cannot belong to its closure. Thus the only possible extra closure point is $a$. The balls at $a$ meet $B$ exactly when arbitrarily small original-metric balls at $a$ do so. ConsequentlyIn the first case $a$ is already included. In the second case a ball smaller than the positive distance separates it from $B$; in the last case every ball meets $B$. If $B$ is empty its closure is empty, with distance convention $d(a,\varnothing )=+\infty$.