Compact subsets of an arbitrary simplicial realization meet finitely many open simplices

If the set $J$ of open simplices meeting $C$ were infinite, for each positive integer $n$ let $X_n$ be the nonempty set of ordered $n$-tuples of points of $C$ with distinct support faces. Countable Choice selects one tuple for each $n$. Flatten these finite tuples into a sequence and retain, in their natural order, the first point in each previously unseen support face. There are infinitely many retained points since the tuple of length $n$ supplies $n$ different faces. This gives a sequence $q_j$ in distinct open simplices, using only the stated countable independent choices and least-index deletions.

A closed simplex $\mathrm{\mid }\tau \mathrm{\mid }$ has finitely many faces. Each retained point in it has a distinct support face contained in $\tau$, so it contains only finitely many $q_j$. Every subset of $Q=\{q_j:j\ge 1\}$ consequently has finite closed trace on every simplex and is weakly closed in $\mathrm{\mid }K\mathrm{\mid }$. Thus $Q$ is closed in $C$ and its subspace topology is discrete. Closedness in compact $C$ makes $Q$ compact, whereas its singleton open cover has no finite subcover. This contradiction proves $J$ finite.

Include all faces of the finitely many simplices in $J$ to obtain a finite subcomplex containing $C$; if $C$ is empty use the vertex-free subcomplex. A continuous image of a compact simplex is compact because any open cover pulls back to an open cover with a finite subcover. A finite union of these images is compact, so the preceding conclusion applies to the entire family at once, including an empty family.