We define an infinite simplicial complex $K$ to be a set of simplices in some $\mathbb R^n$ sch that the following conditions hold:
$1$. Given a simplex in $K$, every "face" of A (i.e., the simplex generated by any subset of the set of vertices of $A$) is also in $K$.
$2$. Given two simplices $A, B\in K$, their intersection is a face of both $A$ and $B$ (or empty).
$3$. The union of the simplices in $K$ is a closed subset of $\mathbb R^n$.
$4$. The topology of $K$, considered as a subspace of $\mathbb R^n$, is the same as the topology of $K$ considered as the identification space of the union of the collection of simplices.
From these definitions, we can show that any point in $K$ has a neighborhood that intersects only finitely many simplices, from which it follows that $K$ is locally compact and locally path-connected.
My professor also claimed that we could show that any infinite complex could only be countably infinite, though our book (Armstrong) does not prove this. Is this statement true, and if so, how to prove it? Obviously, if $K\subseteq\mathbb R^n$, then there can only be countably many $n$-simplices, since each one's interior would contain a point with rational coordinates, of which there are countably many (and the interiors of $n$ simplices can't intersect). But why can't there be uncountably many lower dimensional simplices?
Does it have to do with the fact that any uncountable set of points in $\mathbb R^n$ has a limit point, which would lead to a contradiction to $K$ being closed?