$\DeclareMathOperator\Simp{Simp}\newcommand\geom[1]{\lvert#1\rvert}$I'm not sure if we need a 4th proof of this fact, but after wondering about this for several years I realized it can be proven in a very formulaic manner: turn everything in sight into (the classifying space of) a category and apply Quillen's Theorem A. Let's see how it goes.

The simplicial complex $\geom K$ is more-or-less already a category —
if we think of $K$ as just its set of simplices, then it's partially ordered by inclusion, and the geometric realization of this poset $(K, \subseteq)$ is the usual barycentric subdivision of $\geom K$ (and in particular, it's homeomorphic to $\geom K$).

Next, every simplicial $X$ set can be “turned into a category” by taking the category of simplices $\Simp(X)$. I don't know a simple proof that $\geom{\Simp(X)}$ is homotopy equivalent to $\geom X$, but it is proven in Hirschhorn's book Model Categories and Their Localizations, Theorem 18.9.3.

We want a functor between $\Simp(X(K))$ and $(K, \subseteq)$. This is straightforward; objects of $\Simp(X(K))$ are (in bijection with) simplices of $X(K)$, and we can send a list $(x_0, \dotsc, x_n)$ to the simplex $\{x_0, \dotsc, x_n\}$ in $K$. All diagrams in $K$ commute (since it's a poset), and this makes it easy to verify that this defines a covariant functor $s: \Simp(X(K))\rightarrow (K, \subseteq)$.

Finally, let's try applying Theorem A and see what happens. Fix a simplex $\sigma\in K$. The fiber of $s$ consisting of all simplices in $X(K)$ that map to faces of $\sigma$ can be though of as the category of lists in the set $\sigma$, with a morphism of lists being a way of embedding one list as a sublist of another. It is more-or-less immediate from the definitions that this fiber category is isomorphic to $\Simp(N_* (I(\sigma)))$, where $I (\sigma)$ is the indiscrete category on the set $\sigma$ (that is, the object set of $I(\sigma)$ is $\sigma$, and each morphism set has exactly one element). Since $I(\sigma)$ is equivalent to the trivial category, $\geom{N_* (I(\sigma))}$ is contractible. By the discussion above, so is $\geom{\Simp(N_* (I(\sigma)))}$, and Theorem A says that $\geom s:\, \geom{\Simp X(K)}\stackrel{\simeq}{\rightarrow} \geom K$ is a homotopy equivalence.

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