We define x to be an accumulation point of a set A if every neighbourhood of x contains infinitely many points of A. Let denote the open ball of radius r around x: then a set N is a neighbourhood of x if and only if it contains some open ball of positive radius and in particular of course those open balls themselves are neighbourhoods of x.
So if x is an accumulation point of A then every open ball, being a neighbourhood, contains infinitely many points of A: that is, (i) implies (ii).
An infinite set with one point added or deleted is still infinite, so (ii) is equivalent to (iii).
Now assume (ii). Suppose if possible that x is not an accumulation point of A, so that there is a neighbourhood N of x which contains only finitely many points of A. Since N is a neighbourhood, there is a positive r such that and so and so the latter is finite. This contradicts (ii), and this contradiction shows that (ii) implies (i).