\[T_4\Longrightarrow T_3\Longrightarrow T_2\Longrightarrow T_1\Longrightarrow T_0.\]
Solution
We use the standard conventions: \(T_1\) means all singletons are closed, \(T_2\) means Hausdorff, \(T_3\) means regular \(+T_1\), and \(T_4\) means normal \(+T_1\).
A normal \(T_1\) space is regular: given \(x\notin F\) with \(F\) closed, normality separates the disjoint closed sets \(\{x\}\) and \(F\). Hence \(T_4\Rightarrow T_3\).
If \(X\) is regular \(T_1\) and \(x\ne y\), the set \(\{y\}\) is closed. Choose disjoint open neighborhoods of \(x\) and \(\{y\}\). Hence \(X\) is Hausdorff, so \(T_3\Rightarrow T_2\).
If \(X\) is Hausdorff and \(x\ne y\), there is an open neighborhood of \(x\) not containing \(y\). Thus \[ X\setminus\{y\}=\bigcup_{x\ne y}U_x \] is open, so every singleton is closed. Hence \(T_2\Rightarrow T_1\).
Finally, if \(X\) is \(T_1\) and \(x\ne y\), then \(X\setminus\{y\}\) is an open set containing \(x\) but not \(y\). Thus the two points are topologically distinguishable, so \(T_1\Rightarrow T_0\).
Therefore \[ T_4\Longrightarrow T_3\Longrightarrow T_2\Longrightarrow T_1\Longrightarrow T_0. \]