A finite simplicial complex has a compact Hausdorff realization

If there is no nonempty simplex, the realization is empty and is compact and Hausdorff. Otherwise enumerate the finitely many vertices as $v_1,\dots ,v_N$, with $N\ge 1$. The barycentric coordinate map identifies the realization as a set with $E:={\bigcup }_{\sigma \in K,\sigma \mathrm{\ne }\mathrm{\varnothing }}{E}_{\sigma }\subseteq [0,1{]}^N$, where $E_{\sigma }=\{x\in [0,1{]}^N:{\sum }_{i=1}^N{x}_i=1,x_i=0\text{ if }{v}_i\mathrm{\notin }\sigma \}\mathrm{.}$ On each simplex this identification has precisely its Euclidean simplex topology.

Each $E_{\sigma }$ is closed in ${\mathbb{R}}^N$. Indeed coordinates and their sum are continuous: $\mathrm{\mid }{x}_i-y_i\mathrm{\mid }\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_2$ and $\mathrm{\mid }{\sum }_i{x}_i-{\sum }_i{y}_i\mathrm{\mid }\le N\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$. Thus violating any of its finitely many closed coordinate constraints persists on a sufficiently small ball. The finite union $E$ is closed by [L2] and bounded since it lies in $[0,1{]}^N$.

A subset $F\subseteq E$ is weakly closed exactly when $F\cap E_{\sigma }$ is closed in $E_{\sigma }$ for every nonempty simplex, by taking complements in the weak-open criterion [L1]. If $F$ is weakly closed, each such intersection is closed in ${\mathbb{R}}^N$ by step 2.1 and [L2]. Their finite union is $F$, so $F$ is closed in $E$. Conversely, if $F$ is closed in $E$, all its traces on $E_{\sigma }$ are closed there by [L2], hence $F$ is weakly closed. Therefore the weak topology equals the Euclidean subspace topology on $E$.

By [L3] and step 2.1, $E$ is compact. For two distinct points of $E$, intersect their disjoint Euclidean open balls from [L3] with $E$; these are disjoint subspace-open neighborhoods. Hence $E$ is Hausdorff. Step 3.1 transfers both properties to the realization; step 1.1 handled the empty case.