Partition Topology is T5

Theorem
Let $S$ be a set and let $\mathcal P$ be a partition on $S$ which is not the (trivial) partition of singletons.

Let $T = \left({S, \vartheta}\right)$ be the partition space whose basis is $\mathcal P$.

Then $T$ is a $T_5$ space.

Thus $T$ is also a $T_4$ space.

Proof
Let $A$ and $B$ be subsets of $S$ such that $A^- \cap B = A \cap B^- = \varnothing$.

From Open Sets in Partition Topology are also Closed, we get that $A^-$ and $B^-$ are both clopen.

From Set is Subset of Closure $A \subseteq A^-$.

From $A^- \cap B = \varnothing$ it follows from Intersection of Complement with Subset is Empty that $B \subseteq S \setminus \left({A^-}\right)$.

Since $A^-$ is clopen, it follows that $A^-$ and $S \setminus \left({A^-}\right)$ are both open.

Also, they are disjoint by definition.

Thus we have $A^-$, an open set containing $A$, and $S \setminus \left({A^-}\right)$, an open set containing $B$.

Thus, by definition, $T$ is a $T_5$ space.

The last statement follows from the fact that a $T_5$ space is a $T_4$ space.