I got confused with the definition of a simplicial set and its geometric realization. Is it correct that if I take the following simplicial set then its geometric realization will be a circle?
$X_0=\{a_0,b_0\}$, $X_n=\{a_n\}$ for $0<n$. All the maps are constant with image $a_i$ s.t. $0\leq i$.
What might help you construct examples is these objects that I can't for the life of me remember what they are called. Anyway, of your simplex category $\Delta$ you have a subcategory with the same objects but morphisms are those morphisms which are injective in $\Delta$. This category is called $\Delta_+$.
You have an inclusion $\Delta_+ \rightarrow \Delta$, so a restriction functor $[\Delta^{op},\textbf{Set}] \rightarrow [\Delta_+^{op},\textbf{Set}]$. This restriction functor has a left adjoint $L$ which basically takes a functor $\Delta_+^{op} \rightarrow \textbf{Set}$ and adjoins a bunch of degenerate simplices to make it a simplicial set.
Then if you consider for instance $X:\Delta_+^{op} \rightarrow \textbf{Set}$ given by $X_0=\{p\}$, $X_1=\{s\}$ and $X_n = \emptyset$ else this does indeed define a functor letting morphisms in $\Delta^{op}_+$ get taken to the only possible ones. You should picture $X$ as a circle because you have one $0$-simplex that you attach both ends of a $1$-simplex to. This avoids the problem of having to explicitly define the degenerate simplices.
You can also show that if we define geometric realization of a functor $Y:\Delta_+^{op} \rightarrow \textbf{Set}$ in the same way we define geometric realization of simplicial sets that $\mid Y \mid \cong \mid LY \mid$ for all $Y$, they are homotopy equivalent. This is because $LY$ is just $Y$ with a bunch of degenerate simplices attached so when taking geometric realization you can squish all of these simplices down and end up with a homotopy equivalence. Anyway you can easily show that $\mid X \mid = S^1$ and thus $\mid LX \mid \cong S^1$.
It is not possible to find a simplicial set $S$ and a homeomorphism $\mid S\mid \rightarrow S^1$ since $S$ will always contain degenerate simplices in a high dimension so the best you can do is homotopy equivalence.
Please, if I get too carried away here tell me. I'm happy to explain in a more down to earth manner.