A finite simplicial complex has a compact Hausdorff realization

If $K$ has no nonempty simplices, then $\mathrm{\mid }K\mathrm{\mid }=\mathrm{\varnothing }$, which is compact and Hausdorff. Otherwise $K$ has finitely many vertices; write them as $v_1,\dots ,v_N$. For each simplex $\sigma$ of $K$, the subset $\mathrm{\mid }\sigma \mathrm{\mid }\subseteq \mathrm{\mid }K\mathrm{\mid }$ identifies with a Euclidean simplex in $[0,1{]}^N$, cut out by finitely many linear equations and inequalities, so $\mathrm{\mid }\sigma \mathrm{\mid }$ is compact.

The realization $\mathrm{\mid }K\mathrm{\mid }$ is the union of the finitely many subsets $\mathrm{\mid }\sigma \mathrm{\mid }$ over the nonempty simplices $\sigma$ of $K$, and this union is empty in the case handled at the start of step 1.1. Therefore step 1.1 makes $\mathrm{\mid }K\mathrm{\mid }$ a finite union of compact sets and hence compact.

For each simplex $\sigma$, let $O_{\sigma }\subseteq [0,1{]}^N$ be open with $U\cap \mathrm{\mid }\sigma \mathrm{\mid }=O_{\sigma }\cap \mathrm{\mid }\sigma \mathrm{\mid }$. Since there are only finitely many simplices, a subset $U\subseteq \mathrm{\mid }K\mathrm{\mid }$ is weakly open exactly when $U=({\bigcup }_{\sigma }{O}_{\sigma })\cap \mathrm{\mid }K\mathrm{\mid }$. Thus the weak topology on $\mathrm{\mid }K\mathrm{\mid }$ agrees with the subspace topology from $[0,1{]}^N$. The cube $[0,1{]}^N$ is Hausdorff, so $\mathrm{\mid }K\mathrm{\mid }$ is Hausdorff as a subspace.

Steps 2.1 and 2.2 give compactness and Hausdorffness.