Cantor's finite intersection principle

[Rasalhague]

I'm trying to understand the proof given in the last 10 minutes or so of this video lecture, but after some struggle, it occurs to me that I may be misinterpreting what the theorem says. According to this, Cantor's finite intersection principle states the following.

Given a metric space (X,d) and a collection of compact subsets

\left \{ K_\alpha \subseteq X \; \bigg| \; \alpha \in A \right \}

(where A is an index set), if the elements of any finite subcollection of \left \{ K_\alpha \right \}_{\alpha \in A} have a nonempty intersection, then the intersection of all the K_\alpha is nonempty.

But should "any" be read in the specific sense, "there exists", here:

\left [ \exists \left \{ K_{\alpha_i} \right \}_{i=1}^n \subseteq \left \{ K_\alpha \right \}_{\alpha \in A} \left \left ( \bigcap_{i=1}^{n} K_{\alpha_i} \neq \varnothing \right ) \right ] \Rightarrow \left [ \bigcap_{\alpha \in A} K_\alpha \neq \varnothing \right ] \enspace ?

Or should "any" be read in the nonspecific sense, "for all":

\left [ \forall \left \{ K_{\alpha_i} \right \}_{i=1}^n \subseteq \left \{ K_\alpha \right \}_{\alpha \in A} \left \left ( \bigcap_{i=1}^{n} K_{\alpha_i} \neq \varnothing \right ) \right ] \Rightarrow \left [ \bigcap_{\alpha \in A} K_\alpha \neq \varnothing \right ] \enspace ?

And what sense or senses could collection have here: set, class, family, multiset?

 

[disregardthat]

\left [ \forall \left \{ K_{\alpha_i} \right \}_{i=1}^n \subseteq \left \{ K_\alpha \right \}_{\alpha \in A} \left \left ( \bigcap_{i=1}^{n} K_{\alpha_i} \neq \varnothing \right ) \right ] \Rightarrow \left [ \bigcap_{\alpha \in A} K_\alpha \neq \varnothing \right ] \enspace

It's this,

\forall \left \{ K_{\alpha_i} \right \}_{i=1}^n \subseteq \left \{ K_\alpha \right \}_{\alpha \in A} \left \left ( \bigcap_{i=1}^{n} K_{\alpha_i} \neq \varnothing \right ) \Rightarrow\bigcap_{\alpha \in A} K_\alpha \neq \varnothing \enspace

"For any", "given any" and "for all" are terms used interchangeably. "Collection" means "set" here.

 

[Rasalhague]

Thanks very much, Jarle! No wonder I was getting confused; I'd been taking it in the other sense.