The intersection of $T_1$ topologies. 
A topological space $(X, \tau)$ is a $T_1$ space if $\forall$ $x,y$$\in$$X$, $x$$\neq$$y$ we can find an open set that contains $x$ and not $y$ abd an open set that contains $y$ and not $x$.
A useful proposition about $T_1$ spaces is that in a $T_1$ space, every singleton is a closed set.

Suppose we are given two  $T_1$ spaces $(X,\tau_1)$ and $(X,\tau_2)$.Prove that $(X,\tau_1 \cap\tau_2 )$ is a $T_1$ space.
I did this:
Let $x$$\in$$X$. We know that $\{x\}$ is closed in in $(X,\tau_1)$ and in $(X,\tau_2)$ so $X$\ $\{x\}$ is open in both spaces. Hence $X$\ $\{x\}$$\in$ $\tau_1\cap \tau_2$. It is easily proved that the intersection of topologies in $X$ is again a topology in $X$ and finally we have that $\{x\}$ is closed in $(X,\tau_1 \cap\tau_2 )$.
My general question is that can we prove with the same argument that the intersection of an arbitrary collection of $T_1$ topologies is a $T_1$ topology?
Or is there a tricky counterexample to this last statement?
Thank you in advance!

Each of your topologies contains the co-finite topology ( a set is open if it is the empty set or if its complement is finite.) So the intersection of your topologies contains this topology, and so points are closed in this topology as well.
