Compact implies feebly compact

Statement
Any compact space is a feebly compact space.

Compact space
A topological space is termed compact if every open cover of the space has a finite subcover.

Feebly compact space
A topological space is termed feebly compact if every locally finite collection of nonempty open subsets is finite.

Proof
Given: A compact space $$X$$, a locally finite collection of nonempty open subsets $$V_i, i \in I$$ of $$X$$.

To prove: $$I$$ is finite.

Proof:


 * 1) There exists a point-indexed open cover $$U_x, x \in X$$ of $$X$$ such that each $$U_x$$ intersects finitely many $$V_i$$s: For each point $$x \in X$$, we can find an open subset $$U_x$$ that works by the definition of locally finite. Putting these together, there exists a point-indexed open cover.
 * 2) There exists a finite subset $$A$$ of $$X$$ such that the $$U_a, a \in A$$, cover $$X$$: This follows from the previous step and the definition of compactness.
 * 3) The size of $$I$$ is finite: Every $$V_i,i \in I$$, is a nonempty open subset of $$X$$, hence it intersects at least one of the $$U_a$$s. Thus, to count the $$V_i$$s, it suffices to count the $$V_i$$s that intersect at least one $$U_a$$. For each $$U_a$$, the number of $$V_i$$s intersecting it is finite, and there are finitely many $$a$$s in $$A$$, so the total number is finite.