Proof (sketch)

If $\mathbf{2}$ is Sierpinski space (two points $0$, $1$ and three opens $\emptyset$, $\{1\}$, and $\{0, 1\}$), then the continuous map $I = [0, 1] \to \mathbf{2}$ taking $0$ to $0$ and $t \gt 0$ to $1$ is a weak homotopy equivalence¹.

For any finite topological space $X$ with specialization order $\mathcal{O}(X)$, the topological interval map $I \to \mathbf{2}$ induces a weak homotopy equivalence $B\mathcal{O}(X) \to X$:

(where we implicitly identify $\Delta^{op}$ with the category $Int$ of finite intervals with distinct top and bottom). The isomorphism on the right says that any finite topological space can be constructed by gluing together copies of Sierpinski space, in exactly the same way that any preorder can be constructed by gluing together copies of the preorder $\{0 \leq 1\}$.

On the other hand, any finite simplicial complex $X$ is homotopy equivalent to its barycentric subdivision, which is the geometric realization of the poset of simplices ordered by inclusion. Thus finite posets model the weak homotopy types of finite simplicial complexes.