Theorem:
If $\{K_α\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_α\}$ is nonempty, then $⋂_α K_α$ is nonempty.
What I think is a fake proof:
I've found this proof on the last comment of a post from a concurrent website. I think it is flawed since between $(2)$ and $(3)$ he assumes $X$ is compact...
Here it goes:
Suppose that $$⋂_{α∈A}K_α=∅ \tag{1}$$ then by DeMorgan's law we get that $$⋃_{α∈A}(X\backslash K_α)=X \tag{2}$$ but since the $K_α$ are closed we see that $X\backslash K_α$ are open and so by compactness there exists some finite subcover $X\backslash K_{α_1},⋯,X\backslash K_{α_n}$ but $$⋃_{j=1...n}(X\backslash K_{α_j})=X \tag{3}$$ implies, once again by DeMorgan's law, that $$⋂_{j=1...n}K_{α_j}=∅ \tag{4}$$ contradicting the FIP of $\{K_α\}_{α∈A}$ .
What do you think? Is this proof accurate?
PS: I already have a proof of this result, but I am curious about this alternative proof.