Alexandroff Extension of Rational Number Space is T1 Space

Theorem
Let $\left({\Q, \tau_d}\right)$ be the rational number space under the Euclidean topology $\tau_d$.

Let $p$ be a new element not in $\Q$.

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

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

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

Proof
From Condition for Alexandroff Extension to be $T_1$ Space, $T^*$ is a $T_1$ space $\left({\Q, \tau_d}\right)$ is also a $T_1$ space.

From Rational Numbers form Metric Space, $\left({\Q, d}\right)$ is a metric space.

From Metric Space fulfils all Separation Axioms, $\left({\Q, \tau_d}\right)$ is a $T_1$ space.

Hence the result.