This is question 3b) of exercises on page 14 of Munkres Algebraic Topology.
Show that, in general, if a simplicial complex $K$ is not locally finite, then the space $\vert K \vert$ is not locally compact.
My attempt: I set out to prove this via contradiction. Suppose that a simplicial complex is not locally finite but the space $\vert K \vert$ is locally compact. Since the simplicial complex $K$ is not finite we know there exists a point $v\in \vert K \vert$ which is a member of an infinite number of simplices in $K$. Let these simplices be denoted as $\sigma_{\alpha},\alpha\in J$ where $J$ is some indexing set. Since $\vert K \vert$ is supposed to be locally compact there should exist some neighborhood of $v$ which is compact. This is where I struggle to progress.
I would like to say something like $\cup_{\alpha \in J}\sigma_{\alpha}$ is a cover of this neighborhood and since said neighborhood is compact it has a finite subcover which would contradict the fact that $v$ is supposed to be a member of an infinite number of simplices.
I realise there are so many things wrong with the above logic, first the sets used to make the cover are not open, and thus we can't use the open subcover definition of compactness and I don't even know if the union of all simplices containing $v$ is a cover of the compact neighborhood.
Would really appreciate a hint rather than an answer
Hint: Every compact subset $C$ of $|K|$ can intersect only finitely many simplices, where $K$ is an arbitrary simplicial complex. To prove this, we construct a subset $S$ of $C$ by picking one point from every open simplex intersected by $C$. Now show that $S$ and any of its subsets is closed in $|K|$. This means $S$ is a closed and discrete subset of $C$. Conclude that $S$ must be finite.
By the way, local compactness and local finiteness are actually equivalent. Moreover, they are equivalent to first-countability of $|K|$.