A space is T0, or half Hausdorff, if every pair of points implies an open set containing one and not the other.
If we assume two points x and y violate T0, then the topology cannot distinguish between them. Every open set contains both or neither.
Assume y and z are also inseparable. Suppose x lives in an open set O that does not contain z. If y is not in O then x and y are separable, and if y is in O then y and z are separable. Therefore x and z are also inseparable, and the relation is transitive.
Merge each group of inseparable points into one point. This does not change the topology at all. Thus every space can be crunched down to a space that is half Hausdorff.
A space is T1 if every pair of points x and y has an open set containing x but not y, and an open set containing y but not x. Clearly T1 implies T0.
An equivalent criterion says every point is closed. Given a point p, cover each point in the complement of p with an open set disjoint from p. The union of these open sets is open, and p is closed. Conversely if p is closed its complement is open.
Let a space consist of two points, x and y, with open sets (), (y), and (xy). This space is T0, but not T1.
A space is T2, or Hausdorff, if any two points can be contained in two disjoint open sets. Clearly T2 implies T1.
Consider the "line with two origins". This is a real line, e.g. the x axis, but there are two origins. Open intervals are base sets as usual, but if the interval brackets 0, it must contain exactly one of the two origins. Verify the base criterion for these intervals; the topology is valid. Now, either origin can be placed in an open set apart from the other, yet the origins cannot be separated in two disjoint open sets. The space is T1, but not T2.
A space is T3, or regular, if it is T1, and any closed set and a point not in this set can be contained in two disjoint open sets. Since points are closed (via T1), T3 implies T2.
A space is T4, or normal, if it is T1, and any two disjoint closed sets can be housed in disjoint open sets. Since points are closed (via T1), T4 implies T3.
We're getting a head of ourselves, but let's say you know what a metric space is. Let G and H be disjoint closed sets in a metric space. Consider a point p in the space. The distance from p to G is the greatest lower bound of the distances from p to all the points in G. Since G is closed, p is in G iff its distance to G is 0. Now p also has a distance to H. Both distances cannot be 0, else G and H intersect in p. Assume p is closer to G, by a difference of δ. The points in the ball, centered at p, with radius δ/2, are all closer to G than to H. This is an open set. The set of points that are closer to G than to H is covered by open sets, and is open. This set also contains G. Similarly, the set of points closer to H is open, and contains H. These to open sets are disjoint, hence we have separated G and H. Every metric space is normal.