Topological space $(X,\tau)$ is $T_{1}$ space if for any two points $a,b \in X$ there exist an open neibourhood $U_{a}$ of $a$ with no $b$ in it.
Cofinite topology on set $X$ is $\tau_{cf} = \{ A \in 2^{X} | X \setminus A $ is finite$ \} \cup \{ \emptyset \}$
I need to proof that there is no weaker topology $\tau$ then cofinite topology on an infinite set $X$ that turns $(X,\tau)$ into $T_{1}$ space.
My attempt was to go from the contradiction, that is suppose there is a topology $\tau \subset \tau_{cf}$ that turns $X$ into $T_{1}$ space. $\tau $ consists only of sets from $\tau_{cf}$ and for any two points there exists an open neibourhood of one with no another. But then I stucked. I think I should show that $\tau = \tau_{cf}$.