 Solution

List the finite vertex set as $V=\{v_1,\dots ,v_N\}$. Map a point to its barycentric coordinates:On each simplex this is an affine map and therefore continuous. The definitions agree on every common face, so the quotient topology defining the realization makes $f$ continuous on all of $|K|$. Barycentric coordinates are unique after zero coordinates are added, so $f$ is injective.

Since Euclidean space is Hausdorff, distinct images have disjoint open neighbourhoods. Their inverse images separate the original points. Hence $|K|$ is Hausdorff. A continuous injection into a Hausdorff space is enough for this conclusion; no prior Hausdorff assumption on the domain is being used.