Why is $T_1$ required for a topological space to be $T_4$? Let's say we have some topological space.
Axiom $T_1$ states that for any two points $y \neq x$, there is an open neighborhood $U_y$ of $y$ such that $x \notin U_y$.
Then we say that a topological space is $T_4$ if it is $T_1$ and also satisfies that for any two closed, non-intersecting sets $A,B$, there are open neighborhoods $U_A,U_B$ respectively, such that $U_A\cap U_B = \emptyset$.
Could anyone give an example of a topological space which satisfies the second condition of $T_4$, but which is not $T_1$?

Consider a topological space $X$ with two (different) points and the trivial topology (that is, only $\emptyset$ and $X$ are open sets).
This space satisfies the second condition of $T_4$: the only two closed non-intersecting sets are $\emptyset$ and $X$ and you can take as open neighborhoods $U_\emptyset = \emptyset$ and $U_X = X$.
But $X$ is not $T_1$, as you can easily check.
$T_1$ (also called Fréchet space) is equivalent to the fact that every point is a closed set (exercise: prove this!). So, if you had a regular space ($T_3$) $X$ that was not $T_1$, you could not say that $X$ is $T_2$ (or Hausdorff); that is, you couldn't say that $T_3 \Rightarrow T_2$.
So, it is necessary to add the $T_1$ condition to $T_3$ and $T_4$ in order to have the beautiful sequence of implications that makes all of us happy :-)
$$
T_4 \ \text{(normal)} \ \Longrightarrow \
T_3 \ \text{(regular)} \ \Longrightarrow \
T_2 \ \text{(Hausdorff)} \ \Longrightarrow \
T_1 \ \text{(Fréchet)}
$$
