First of all, go read metric space. Each metric space is a set of points X combined with a distance rule d: X x X -> R following particular rules.

Let P be the set of real numbers greater than . For an x ∈ X and i ∈> P, Ni(x) (the "i-neighborhood of x") is the set of points y such that d (x, y) < i. (Imagine all of the possible spheres in space, minus their surfaces).

Suppose i > 0, y ∈ Ni(x) and y != x. Notice that d (x, y) > 0, otherwise d would not be a metric. And d (x, y) < i from the definition of Ni(x). Thus, 0 < d (x, y) < i.

So, given a point y ∈ Ni(x) such that y != x, some i' must exist, 0 < i' < i,  for which Ni'(y) ⊂ Ni(x).

Because of that, for all x1, x2 ∈ X and i1, i2 ∈ P, the intersection of Ni1(x1) and Ni2(x2) (if not empty) must be the union of several Ni'(y). (This part isn't necessary for the proof, and doesn't appear in actual descriptions, but I included it because it was the link I needed to understand it all). Briefly put, it's turtles all the way down.

Because of that, the collection of all Ni(x) for every x X and i P can be used as a base for a topology of X.

This topology is the called the space's "metric topology".  It allows us to describe the metric space in terms that apply to topological spaces.