Closed good cover of a triangulable space By a good closed cover of a topological space $X$, I mean a collection of closed subspaces of $X$, such that the interior of them cover $X$, and any finite intersection of these closed subspaces is contractible.
Every triangulable space $X$ admits a good open cover: just fix a triangulation and take open stars at all vertices. As for closed stars, the interior of closed stars cover $X$, but the intersection of closed stars can be non-contractible, as the simple example of the circle  (3 vertices, 3 segments) shows: the intersection of two different closed stars is the disjoint union of a segment and a vertex. However, after taking barycentric subdivision, we get 6 vertices and 6 segments in the circle, and the closed stars now form a good closed cover.
My question is: is it true that after iterated barycentric subdivisions, the closed stars of a triangulated space will form a good closed cover? (If not, is it true for finite / locally finite geometric simplicial complexes?)
Thank you!
 A: Claim: Let $Z$ be a simplicial complex. For each simplex $\sigma\in Z$, let $N_2 (\sigma, Z)$
denote the simplicial neighborhood of $\sigma$ (or really, the second barycentric subdivision of $\sigma$)
inside the second barycentric subdivision of $Z$. Then
$\{|N_2 (\sigma, Z)|: \sigma \in Z\}$ is a good cover of $|Z|$.
Before discussing the proof, let's fix terminology and notation. Here a simplicial complex will mean a set $K$ such that each element $\sigma\in K$ is a non-empty finite set, and such that $\sigma'\subseteq \sigma \in K$ implies $\sigma'\in K$ (unless $\sigma' = \emptyset$). The geometric realization of a simplicial complex will be denoted $|K|$.
The subdivision of a simplicial complex $K$ is the simplicial complex $sd (K)$ whose elements are the non-empty chains (totally ordered subsets) of $K$. (Here we are implicitly viewing $K$ as a poset under set-theoretic inclusion; this will keep happening.) Iterating the barycentric subdivision functor $n$ times gives $sd^n (K)$ (and $sd^0 K := K$). It will be useful to consider the function $min: sd^n (K) \rightarrow sd^{n-1} K$, which sends a chain to its minimum element. This is an order-reversing function (but not a simplicial map). We'll also use the composite function $min^2 = min\circ min : sd^2 K\to K$; this is order-preserving.
If $K \subseteq Z$ is a subcomplex, the simplicial neighborhood of $K$ in $Z$ is defined by
$N(K, Z) := \{\sigma\in Z : \exists \tau \in Z, \sigma \subseteq \tau \supseteq \kappa \in K\}.$
If $K$ is a subcomplex of $Z$, then $sd^m K$ is a subcomplex of $sd^m Z$, and we simplify notation by defining
$N_m (K, Z) := N(sd^m K, sd^m Z).$
The arguments below are all very simple, but can be quite confusing to read because we have to keep track of things like the difference between a simplex $\sigma \in K$, a 1-element chain $\{\sigma\} \in sd K$, and a 1-element chain $\{\{\sigma\}\} \in sd^2 K$ whose single element happens to itself be a 1-element chain.
I recommend thinking of elements $C \in sd^2 Z$ as 2-dimensional arrays, where each column is a chain of simplices in $Z$ and moving from left to right, we simply add more elements to the chain.
Lemma 1:  For any subcomplex $K$ of a simplicial complex $Z$, we have
$N_2 (K, Z) = \{ C \in sd^2 Z : min^2 C \in K\}.$
Moreover, this simplicial complex is precisely the order complex of the subposet $P_1 (K, Z)\subseteq sd Z$ defined by
$P_1 (K, Z) := \{c \in sd Z : min(c) \in K\}.$
Note: By the order complex of a poset $P$, I just mean the set of (finite) non-empty chains in $P$. Note here that $P_1 (K, Z)$ is not a simplicial complex (in general), so in particular it's not a subcomplex of $Z$. But for each subset $P \subseteq Z$, the  set of chains in $P$ is a subcomplex of $sd Z$.
Proof: We prove containment in both directions. Say $C\in N_2 (K, Z)$. Then we have $C \subseteq D \supseteq E$
for some $D \in sd^2 Z$ and $E \in sd^2 K$. Since $min$ is order-reversing, $D \supseteq E$ implies $min(D) \subseteq min(E)$, and since $min(E) \in sd K$ we have $min(D) \in sd K$ as well. Similarly,
$C\subseteq D \Longrightarrow   min^2 C \subseteq min^2 D$.
But $min(D) \in sd K$ implies $min^2 D \in K$, so $min^2 C\in K$ as well.
The reverse inclusion is simpler. Say $C \in sd^2 Z$ and $min^2 C \in K$. Let $C_0' = \{min^2 C\} \in sd K$, and note that $C_0' \subseteq min(C)$, so $\{ C_0'\}\cup C\in sd^2 Z$. We now have
$C \subseteq \{ C_0'\}\cup C  \supseteq   \{ C_0' \}
\in sd^2 K,$
showing that $C\in N_2 (K, Z)$.
Finally, note that if $C\in sd^2 Z$ and $min^2 C \in K$, then in fact
$min (c) \in K$ for all $c\in C$, because for each $c\in C$,
$min(C) \subseteq c \implies min(c) \subseteq min^2 C \in K.$
Thus $N_2 (K, Z)$ is the order complex of the poset $\{c \in sd Z :  min(c) \in K\}$. QED
Here is an immediate but useful consequence of the above characterization of neighborhoods:
Lemma 2: If $Z$ is a simplicial complex and $K, L\subseteq Z$ are subcomplexes, then $N_2(K, Z) \cap N_2 (L, Z) = N_2 (K\cap L, Z)$.
The following is classical and well-known, although I don't know where to find an elementary proof in the literature.
Lemma 3: For any subcomplex $K$ of a simplicial complex $Z$, the inclusion $sd^2 K \hookrightarrow N_2 (K, Z)$ is a homotopy equivalence.
Proof: It is sufficient to check that the inclusion $i : sd K \hookrightarrow P_1 (K,Z)$ satisfies the hypotheses of Quillen's Fiber Theorem (aka Theorem A) - that is, it suffices to show that the Quillen fibers are all contractible.
For each $c\in P_1 (K, Z)$, the Quillen fiber "under" $c$ is just $\{d \in sd K : d \subseteq c\}$, and has $c\cap K$ as its maximum element (note that $c\cap K$ is non-empty, as $c\in P_1 (K, Z)$ implies $\min (c)\in c\cap K$). Every poset with a maximum element has contractible realization. QED
Aside: This argument actually leads to the stronger conclusion that $K \hookrightarrow N_2 (K, Z)$ is a simple homotopy equivalence, by a result of Barmak in On Quillen's Theorem A for posets (arXiv, J. Comb. Theory Ser. A).
Now we can prove the Claim regarding good covers:
Proof of Claim: For each simplex $\sigma\in Z$, there is a corresponding subcomplex of $Z$, which we also denote by $\sigma$, consisting of all non-empty subsets of $\sigma$. Since $|\sigma|$ is contractible, Lemma 3 implies that $|N_2 (\sigma, Z)|$ is contractible as well. Moreover, Lemma 2 (and induction) shows that every finite intersection $\bigcap_i |N_2 (\sigma_i, Z)|$ has the form $|N_2 (\bigcap_i \sigma_i, Z)|$, and  $\bigcap_i \sigma_i$ is either empty or is just (the subcomplex corresponding to) another simplex. So $\bigcap_i |N_2 (\sigma_i, Z)|$ is either empty or contractible. QED
