Why is the the following statement not equivalent to compactness?

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Well it comes down to word-play again. I'm confused to the core of my bones as to why the following isn't equivalent to saying that a space is compact

Every open cover is finite.

A compact set must have every open cover in which there is a finite subcover.

Well, the statement tells me that every single possible open cover is finite. So...why doesn't this qualify?

Subcover needn't be a proper subset/cover so if I have every open cover being finite...the open cover itself can act as a subcover(which is of course finite) so voila, we're done..no?

Elaborate explanation much appreciated on this

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A compact space is one in which every open cover has a finite subcover. Not every subcover has to be finite. So, for example, the cover $C = \{(-\varepsilon,\varepsilon)\,|\, 0 < \varepsilon\}$ is an open cover of $[-1,1]$, and has as a finite subcover the singleton $\{(-2,2)\}$. The cover is still infinite, however.