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I'll expand upon my comment in Andre's answer. In some sense (which I am about to make precise), non-Hausdorff spaces occur when trying to "naturally" close Hausdorff spaces under colimits. Let's say the only spaces you think are "real" are compact Hausdorff spaces (this is somewhat reasonable, from certain viewpoints). But then, you might want to consider an infinite disjoint union of such spaces as still being a space, so you arrive at having to consider locally compact Hausdorff spaces. In fact, EVERY compactly generated space (not assuming any separation axioms) is the quotient of a (possibly) infinite disjoint union of compact Hausdorff spaces.

To see this: Any compactly generated space $X$ is a (possibly large) colimit of compact Hausdorff spaces. Consider the set $P(X)\O(X)$ of non-open subsets of $X$. Then for element $V$, there exists a map $p_V:T_V \to X$ form from a compact Hausdorff space such that $p^{-1}\left(V\right)$ is not open. Now, the colimit of the diagram $\left(p_V:T_V \to X\right)$ is ALMOST $X$. We just need to make sure The colimit is ALMOST formed by taking a quotient of the disjoint union of each of these $T_V$s- this is true once we know that all points of $X$ are hit, but, we can fix this by adding in a bunch of constant maps to this family.

The converse, that the quotient of a sum of compact Hausdorff spaces is compactly generated is clear.

So, Andre's answer is the "total" answer, in that it includes all (compactly generated) spaces. So yes, (almost) every example of a non-Hausdorff space is really just considering points to actually be subsets of a particular Hausdorff one.

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I'll expand upon my comment in Andre's answer. In some sense (which I am about to make precise), non-Hausdorff spaces occur when trying to "naturally" close Hausdorff spaces under colimits. Let's say the only spaces you think are "real" are compact Hausdorff spaces (this is somewhat reasonable, from certain viewpoints). But then, you might want to consider an infinite disjoint union of such spaces as still being a space, so you arrive at having to consider locally compact Hausdorff spaces. In fact, EVERY compactly generated space (not assuming any separation axioms) is the quotient of a (possibly) infinite disjoint union of compact Hausdorff spaces.

To see this: Any compactly generated space $X$ is a (possibly large) colimit of compact Hausdorff spaces. Consider the set $P(X)\O(X)$ of non-open subsets of $X$. Then for element $V$, there exists a map $p_V:T_V \to X$ form a compact Hausdorff space such that $p^{-1}\left(V\right)$ is not open. Now, the colimit of the diagram $\left(p_V:T_V \to X\right)$ is ALMOST $X$. We just need to make sure all points of $X$ are hit, but, we can fix this by adding in a bunch of constant maps to this family.

The converse, that the quotient of a sum of compact Hausdorff spaces is compactly generated is clear.

So, Andre's answer is the "total" answer, in that it includes all (compactly generated) spaces. So yes, (almost) every example of a non-Hausdorff space is really just considering points to actually be subsets of a particular Hausdorff one.