I'm having trouble understanding some basic facts on algebraic topology. If I have a topological space $X$ and want to calculate the (co)homology of it, it seems to me to calculate singular (co)homology. But this it's not so simple in practice, because (in general) I can't find a nice basis for $S_n(X)$ (set of singular $n$-simplex of $X$). I read about geometric realization in books and found here (in mathstack) some phrases like: "if you want to say something about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$ (i.e. you triangulate $X$ and find the homology groups of the resulting simplicial complex)." Since homeomorphic spaces have the same homology, this solves the problem to calculate singular (co)homology.
So, in the specific case of partially ordered sets (posets) I found a lot of references in which the authors calculate (co)homology of posets using order complex * and geometric realization **. But no one shows that the realization is homeomorphic to the poset (could be because it's obvious). This is not so obvious to me, for example:
Consider the poset $P=\{1,2,3 | 1\leq 2, 1\leq 3\}$. As a topological space we have the open sets $\{\{1,2,3\}, \{2\}, \{3\}, \emptyset\}$ and closed sets $\{\{1\}, \{1,2\}, \{1,3\}\}$. So the order poset $\Delta(P)=\{\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}\}$ and the geometric realization of it $|\Delta(P)|=\{[1],[2],[3],[1,2],[1,3]\}$. I know that must be very dumb, but I can't figure out how $|\Delta(P)|$ is homeomorphic to $P$ (they have the same shape, but how to explicitly describe the homeomorphism?).
Question: How can I show that the geometric realization $|\Delta(P)|$ is homeomorphic to $P$ (in general)?
*Order complex: To every poset $P$, one can associate an abstract simplicial complex $\Delta(P)$ called the order complex of $P$. The vertices of $\Delta(P)$ are the elements of $P$ and the faces of $\Delta(P)$ are the chains (i.e., totally ordered subsets) of $P$.
**Geometric realization: From a geometric simplicial complex $K$ in $\mathbb{R}^n$, one gets an abstract simplicial complex $\Delta(K)$ by letting the faces of $\Delta(K)$ be the vertex sets of the simplices of $K$. For every abstract simplicial complex $\Delta$ there exists a geometric simplicial complex $K$ such that $\Delta=\Delta(K)$. Taking the union of the simplices of $K$ under the usual topology on $\mathbb{R}^n$, is refer to the geometric realization of $\Delta$.