Question regarding a compact class of sets I am currently reading Aliprantis and Border (Third Edition) on the Kolmogorov Extension Theorem, and I seem to be missing a step in the proof that is being claimed. First, a family $\mathbb{K} \subset X$ is a compact class if every countable collection of $\mathbb{K}$ satisfying the finite intersection property has non-empty intersection.
Now, we have Lemma 15.24, which is left unproven:
Lemma 15.24: Let $\mathbb{K} \subset X$ be a compact class. Then the smallest family of sets containing $\mathbb{K}$ that is closed under finite unions and countable intersections is also a compact class.
Fine. But then in the next Lemma, it states that if we allow $\mathbb{K}_u$ to be the collection of all finite unions of a compact class $\mathbb{K}$, then by Lemma 1, it is also a compact class. However, this seems to be ignoring the countable intersection requirement. Am I missing something?
Mark
 A: The definition of compact in this context has the property that subsets of compact classes are also compact (provided they are non-empty.)

However, a more direct way of showing $\mathbb{K}_u$ is compact would be to show that for every $\mathcal{F} = \{X_n: n \in \mathbb{N}\} \subset \mathbb{K}_u$ having the finite intersection property, the intersection $I =\cap \mathcal{F}$ is non-empty.
To that end, by definition, given $n\in\mathbb{N}$, we can find a finite subset $H_n = \{Y^n_{k}: k\le M_n\} \subset \mathbb{K}$ with the property $X_n = \cup H_n = \bigcup_{k\le M_n} Y^n_{k}$. Next, noting that $\mathcal{F}$ has the finite intersection property, we can conclude that 
$$ F_n = \cap \{X_k: k\le n \} = \bigcap_{k \le n} \bigcup_{i \le M_k} Y^k_{i} \neq\emptyset.$$
With that we are ready for the fun bit, 
Claim: For each $n\in\mathbb{N}$ and function $\sigma : [0,n] \rightarrow \mathbb{N}$, with $\sigma(k) \le M_k$, we have 
$$ Y(\sigma) = \bigcap \{ Y^k_{\sigma(k)}: k \le n \} \subset F_n. $$
Moreover, for every $x \in F_n$ there is some $\sigma:[0,n] \rightarrow \mathbb{N}$ with $x \in Y(\sigma)$.
Proof: For $n=0$, the claim follows by definition, since $F_0=X_0 = \cup \{Y^0_i: i\le M_i\}$. So assume we have established the claim for all $k\le n$; then applying the distributive laws for union and intersection produces 
$$F_{n+1} = X_{n+1} \cap F_n = F_n \cap \cup\{Y^{n+1}_k: k \le M_{n+1} \} =  \cup\{F_n \cap Y^{n+1}_k: k \le M_{n+1} \}.$$
It follows that for each $x \in F_{n+1}$ there is some $k \le M_{n+1}$ with $x \in F_n \cap Y^{n+1}_k$; so by the induction hypothesis, we can find a function $\sigma_0: [0,n] \rightarrow \mathbb{N}$ with $x \in Y(\sigma_0) \subset F_n$. Letting $\sigma = \sigma_0 \cup \{(n+1, k)\}$ defines a function mapping $[0, n+1]$ into $\mathbb{N}$ such that 
$$x \in Y(\sigma) = Y(\sigma_0) \cap Y^{n+1}_k \subset F_{n+1}$$
establishing the claim. $\square$
To finish things off, we apply König's lemma to the graph $G=(V,E)$ whose vertex set $V$ consists of functions $\sigma$ for which $Y(\sigma)$ is defined and non-empty. Moreover, we'll assert that $G$ has an edge connecting the vertices $\sigma, \tau\in V$ precisely when $\tau = \sigma \cup \{(n,k)\}$ and $n=\vert\tau\vert$. 
Then $G$ meets the hypothesis of König's lemma and so has an infinite path $\{ \sigma_n : n\in\mathbb{N} \}\subset V$. 
Setting $f = \cup \{ \sigma_n : n \in \mathbb{N} \}$, defines a function taking $\mathbb{N}$ into $\mathbb{N}$ with the property that 
$$ \{ Y^n_{f(n)}: n \in \mathbb{N} \} \subset \mathbb{K} $$ 
has the finite intersection property. Since $\mathbb{K}$ is compact, the intersection $J = \cap \{ Y^n_{f(n)}: n \in \mathbb{N} \} \subset I$ is non-empty. 
