Either-Or Topology is Non-Meager/Proof 1

Theorem
Let $T = \left({S, \tau}\right)$ be the either-or space.

Then $T$ is a second category space.

Proof
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$.

So every non-empty subset of $T$ (apart from $\left\{{0}\right\})$ contains at least one open set of $T$.

So no subset of $T$ is nowhere dense in $T$.

So $T$ can not be the union of a countable set of subsets of $X$ which are nowhere dense in $X$.

Hence the result by definition of second category space.