An exercise concerning simplicial complexes The following exercise is drawn from Ch.24 of Fulton's "Algebraic Topology: A First Course."
Exercise 24.37 (a) and (b)
(a) If $U = \{U_v: v \in V\}$ is a finite collection of open sets whose union is a space $X$, define a simplicial complex, called the nerve of $U$ and denoted $N(U)$, by taking $V$ to be the set of vertices, and defining the simplices to be the subsets $S$ of $U$ such that the intersection of the $U_v$ for $v \in S$ is nonempty. Show that, in fact, $N(U)$ is a simplicial complex.
(b) In turn, if $K$ is any simplicial complex, and $v$ is a vertex of $K$, define an open set $St(v)$ in $|K|$, called the star of $v$, to be the union of the "interiors" of the simplices that contain $v$, i.e., $St(v)$ is the complement in $|K|$ of the union of those $|\sigma|$ for which $\sigma$ does not contain $v$. Show that the open sets $\{St(v): v\in V \}$ form an open covering of $|K|$, and that the nerve of this covering is the same as $K$.
For (a), I want to verify that $N(U)$ satisfies the definition of abstract simplicial complex, because I would think the desired claim follows from there. I think that verifying this will not be too hard, but I don't know if this strategy is appropriate, and wanted to see if anyone visiting would approach the problem differently.
As for (b), I don't have as well-defined a strategy in mind, and wanted to see how anyone visiting might approach the problem. 
 A: You have the right idea for (a). It follows directly from the definition of an abstract simplicial complex. You want to show that if $\sigma \in N(U)$ and $\tau \subset \sigma$, then $\tau \in N(U)$. I.e.,
$$
\bigcap_{v \in \sigma} U_v \neq \emptyset \Rightarrow \bigcap_{v \in \tau} U_v \neq \emptyset.
$$
For part (b) it also pays to consider the definitions carefully. Specifically the definition of $|K|$. I found Fulton's book in Google Books online. Fulton has made a simplifying assumption which makes this problem easier than might be guessed from the information given in your post: Simplicial complexes are defined to have a finite vertex set $V$. In this case $|K|$ has been defined as a set in $\mathbb{R}^{|V|}$.
So,  if $W = \{\text{St}(v) \, | \, v \in V \}$, there are three things to show: (i) $W$ is an open cover of $|K|$, (ii) $K \subseteq N(W)$, and (iii) $N(W) \subseteq K$, but all three things come down to the same essential observation.
(i) You need to verify that $\text{St}(v)$ is open in $|K|$. Use the definitions. Then it remains to show that the union of the $\text{St}(v)$ is $|K|$.
If $x \in |K|$, consider the set $\{ \sigma \in K \, | \, x \in |\sigma| \}$. Take $\tau$ from this set such that $\tau$ has the smallest dimension (i.e., fewest vertices). Now this is the main exercise: Show that for any $v \in \tau$, we have $x \in \text{St}(v)$. It follows then that $W$ is a cover for $|K|$.
(ii) To show $K \subseteq N(W)$, choose a $\sigma \in K$ and consider $x \in \text{int}|\sigma|$. By the same argument needed to complete (i), you can show that $\sigma \in N(W)$.
(iii) Conversely, to show $N(W) \subseteq K$, choose $\sigma \in N(W)$ and note that by definition there exists an $x \in \cap_{v \in \sigma} \text{St}(v)$. [EDIT] Show that this implies $x \in |\tau|$, for some $\tau$ with $\sigma \subseteq \tau$, 
and therefore $\tau \in K$ by the definition of $|K|$. Since $K$ is a complex, we also have $\sigma \in K$.
