Double Pointed Topology is not T0/Proof 1

Proof
Let $T_1 = \left({S, \tau_S}\right)$ be a topological space.

By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$.

Let $x \in S$, and consider the point $\left({x, a}\right) \in S \times A$.

Then:
 * $\forall U \in \tau: \left({x, a}\right) \in U \implies \left({x, b}\right) \in U$
 * $\forall U \in \tau: \left({x, b}\right) \in U \implies \left({x, a}\right) \in U$

as $T_2$ is an indiscrete space.

Hence the result, by definition of $T_0$ (Kolmogorov) space.