Check $T_1$: in $T_1$, every singleton must be closed. $\{a\}$ is open, so $\{b,c\}$ is closed. $\{b\}$ is open, so $\{a,c\}$ is closed. But $\{c\}$: its complement $\{a,b\}\in\tau$ is open, so $\{c\}$ is closed. So singletons are closed → $T_1$ holds. Check $T_2$ (Hausdorff): for $a$ and $c$, we need disjoint open sets. The open sets containing $c$ are only $X$, and any open set containing $a$ is $\{a\}$ or $\{a,b\}$ or $X$ — none disjoint from $X$. Not $T_2$. So $X$ is $T_1$.