Let $X$ be a set. Show that the collection $$T_\infty=\{U : X-U \text{ is infinite or empty or all of $X$} \}$$  is a topology on $X$.  Well $X-X=\emptyset$, so $X \in T_\infty$. And $X-\emptyset= X$ and therefore in $T_\infty$.   Let assume that $\{U_\alpha\}$ is indexed family of sets of $T_\alpha$ that are not empty or $X$. Than  $$X-\cup U_\alpha =\cap X- U_\alpha$$ it seems strange to me that this intersection must be infinite. I don't know $(-\infty,0]$ and $[0,\infty)$ have a finite intersection. Maybe I oversee something.
Can someone please verify my proof?     Is the collection $\tau_\infty = \{U:X-U$ is infinite or empty or all of $X\}$ a topology on $X$?   No. Let $X = \mathbb{R}$.  Clearly, $\{x\} \in \tau_\infty$ for all $x \in X$.  Set $\displaystyle{B = \cup_{x \neq 0} \{x\}}$. Clearly, $X - B = \{0\}$, but $\{0\} \notin \tau_\infty$. Since $\tau_\infty$ is not closed under arbitrary unions, it does not define a topology on $X$.
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