Why is $T_1$ required for a topological space to be $T_4$?

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 :-)

$$ T_4 \ \text{(normal)} \ \Longrightarrow \ T_3 \ \text{(regular)} \ \Longrightarrow \ T_2 \ \text{(Hausdorff)} \ \Longrightarrow \ T_1 \ \text{(Fréchet)} $$

As an addition: the separating closed disjoint sets part is often called normality (a space is normal if it satisfies this), so $T_4$ is normal plus $T_1$, and similarly for regular: $T_3$ is regular and $T_1$. Spaces that have no disjoint non-empty closed sets (besides the trivial topology, we have examples like $\mathbf{N}$ with the topology generated by the sets of the form $U(n) = \{ k : k \ge n \}$ e.g.) trivially satisfy normality. $T_4$ is to avoid these pathologies: the extra $T_1$ ensures that at least all finite sets are closed, so we have some "relevant" closed sets to apply normality to...