T1 not implies Hausdorff

This article gives the statement and possibly, proof, of a non-implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., T1 space (?)) need not satisfy the second topological space property (i.e., Hausdorff space (?))
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Statement

A T1 space need not be a Hausdorff space.

Proof

Example of the cofinite topology

Consider a countable set, say the set $\{1,2,3,\dots \}$ of natural numbers, equipped with the cofinite topology. In the cofinite topology, the nonempty open subsets are precisely the cofinite subsets -- the subsets whose complement is finite. With this topology:

• The space is $T_{1}$: Every point is closed, because its complement is an open subset. Equivalently, if $x,y$ are two natural numbers, the complement of $x$ is an open subset containing $y$ but not $x$.
• The space is not Hausdorff: it is not possible to separate two points with disjoint open subsets, because any two nonempty open subsets have an infinite intersection. This is because the union of their complements is a union of two finite subsets, and hence is not the whole space.