Condition for Alexandroff Extension to be T1 Space

Theorem
Let $T = \left({S, \tau}\right)$ be a non-empty topological space.

Let $p$ be a new element not in $S$.

Let $S^* := S \cup \left\{{p}\right\}$.

Let $T^* = \left({S^*, \tau^*}\right)$ be the Alexandroff extension on $S$.

Then $T^*$ is a $T_1$ (Fréchet) space $T$ is a $T_1$ (Fréchet) space.

Necessary Condition
Let $T = \left({S, \tau}\right)$ be a $T_1$ space.

By definition, $T$ is a $T_1$ space all points of $S$ are closed in $T$.

We have that $S$ is open in $T$ by definition of a topology.

Thus by definition of the Alexandroff extension, $S$ is open in $T^*$.

So as $S = S^* \setminus \left\{ {p}\right\}$ is open in $T^*$, it follows that $\left\{ {p}\right\}$ is closed in $T^*$.

That is, $p$ is a closed point of $T^*$.

Now let $x \in S^*$ such that $x \ne p$.

As $T$ is a $T_1$ space, $\left\{ {x}\right\}$ is closed in $T$.

From Finite Topological Space is Compact, $\left\{ {x}\right\}$ is a compact subset of $T$.

Thus by definition of the Alexandroff extension, $S^* \setminus \left\{ {x}\right\}$ is open in $T^*$.

Thus by definition $\left\{ {x}\right\}$ is closed in $T^*$.

That is, $x$ is a closed point of $T^*$.

Thus it has been shown that all points in $T^*$ are closed points of $T^*$.

Thus by definition $T^*$ is a $T_1$ space.

Sufficient Condition
Let $T^*$ be a $T_1$ space.

By definition of $T$ is a subspace of $T^*$.

By $T_1$ Property is Hereditary, it follows that $T$ is a $T_1$ space.