An interesting topological space with $4$ elements There is an interesting topological space $X$ with just four elements $\eta,\eta',x,x'$ whose nontrivial open subsets are $\{\eta\},\{\eta'\},\{\eta,\eta'\}, \{\eta,x,\eta'\}, \{\eta,x',\eta'\}$. This seems to be some "discrete model" for the $1$-sphere $S^1 \subseteq \mathbb{C}$: The open sets $\{\eta,x,\eta'\}$ and $\{\eta,x',\eta'\}$ may be imagined as arcs joining $\eta$ and $\eta'$ via $x$ or resp. $x'$. They are contractible, and their intersection is the discrete space $\{\eta\} \sqcup \{\eta'\}$. It also follows that $\pi_1(X) \cong \mathbb{Z}$. Without knowing any algebraic topology, one can explicitly classify all coverings of $X$, namely this category is equialent to $\mathbb{Z}\mathsf{-Sets}$.
In contrast to $S^1$, actually $X$ is homeomorphic to the spectrum of a ring: Let $R$ be the localization of $\mathbb{Z}$ at all primes $p \neq 2,3$. There is a canonical surjective homomorphism $R \to \mathbb{Z}/6$. Let $A$ be the fiber product $R \times_{\mathbb{Z}/6} R$. Then $X \cong |\mathrm{Spec}(A)|$. We glue $\mathrm{Spec}(R) = \{\eta,x,x'\}$ with itsself along its closed subscheme $\{x,x'\}$. We end up with two generic points $\eta,\eta'$.
Question. Does this space $X$ have a name? What is the precise relationship to $S^1$? Where do the observations above appear in the literature?
According to Miha's comment below (which is an answer), $X$ is called the pseudocircle, and the definite source for the general phenomenon is:
Singular homology groups and homotopy groups of finite topological spaces, by Michael C. McCord, Duke Math. J., 33(1966), 465-474, doi:10.1215/S0012-7094-66-03352-7.
The pseudocircle already appeared a couple of times on math.SE, for example in question/56500.
 A: Quoting the actual answer compiled by the asker from comments to remove the question from unanswered pool.

According to Miha's comment below (which is an answer), $X$ is called the pseudocircle, and the definite source for the general phenomenon is:
Singular homology groups and homotopy groups of finite topological spaces, by Michael C. McCord, Duke Math. J., 33(1966), 465-474, doi:10.1215/S0012-7094-66-03352-7.
The pseudocircle already appeared a couple of times on math.SE, for example in question/56500.

A: The natural relationship between $S^1$ and $X$ arises as follows.  Consider $X$ as a preordered set in its specialization order: explicitly, the order is $\eta,\eta'\leq x,x'$ (or the reverse of that depending on your conventions).  We can then take the nerve $N(X)$: explicitly, $N(X)$ is a simplicial set in which an $n$-simplex is an order-preserving map $[n]\to X$.  The nondegenerate simplices are the injective order-preserving maps, i.e. the chains in the poset $X$.  There are four nondegenerate $0$-simplices (the $4$ points of $x$) and $4$ nondegenerate $1$-chains (the $4$ chains in $X$ of size $2$), which connect the points of $X$ cyclically: $\eta\leq x$, then $x\geq \eta'$, then $\eta'\leq x'$, then $x'\geq \eta'$.
This means that the geometric realization $|N(X)|$ is homeomorphic to a circle (it's just a cyclic graph with 4 vertices).  Moreover, there is a canonical continuous map $p:|N(X)|\to X$, which sends a point that is in the interior of a nondegenerate simplex $\sigma$ (i.e., a chain in $X$) to its least element in $X$.  Explicitly, this maps the four vertices of $|N(X)|$ to the corresponding points of $X$, and the interiors of the edges to either $\eta$ or $\eta'$ depending on which of the two they have as a vertex.
Now for the magical fact: this canonical map $p:|N(X)|\to X$ is a weak homotopy equivalence, so $X$ is weak homotopy equivalent to $S^1$.  And even more magically, this generalizes.  In particular, McCord proved the following results in his paper Singular homology groups and homotopy groups of finite topological spaces:

Theorem: Let $P$ be any preordered set, considered as a topological space by letting the open sets be the downward closed sets (i.e., the finest topology with $P$ as its specialization order).  Then the natural map $|N(P)|\to P$ constructed as above is a weak homotopy equivalence.
Corollary: Let $X$ be any finite topological space and equip it with its specialization order.  Then the natural map $|N(X)|\to X$ is a weak homotopy equivalence.
Proof: A topology on a finite set is determined by its specialization order, so the topology on $X$ is the same as the topology induced by the specialization order as in the Theorem.
Corollary: Every finite CW-complex is weak homotopy equivalent to a finite topological space.
Proof:  Let $X$ be a finite CW-complex.  Then $X$ is homotopy equivalent to the geometric realization of some finite simplicial complex $Y$.  Let $P$ be the poset of faces of $Y$, ordered by inclusion.  Then $N(P)$ is just the barycentric subdivision of $Y$, and in particular there is a natural homeomorphism $|N(P)|\to |Y|$.  Thus by the Theorem, $X$ is weak homotopy equivalent to $P$.

Note in particular that these results show that the relationship between your space $X$ and $S^1$ is "one-sided": $S^1$ is naturally associated to $X$, but $X$ is not naturally associated to $S^1$.  Indeed, there are many different "finite models" of $S^1$; for instance, the poset of faces in any triangulation of $S^1$ is a "finite model" of $S^1$ as in the proof of the second Corollary.
