Alexandroff Extension of Rational Number Space is Sequentially Compact

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 sequentially compact space.

Proof
The strategy here is to demonstrate that every sequence in $T^*$ is either contained in a compact subspace of $T^*$, or must contain a subsequence which converges to $p$.