Is the Heine Borel property equivalent to countable compactness?

Their definitions seem to be very similar, what is the difference?

Here are the definitions I'm using:

Heine-Borel: Every open cover for $$E$$ admits a finite subcover.

Countable compactness: Let $$(G_i)^{\infty}_1$$ be a sequence of open sets in metric space $$M$$. Then $$E \subseteq M$$ is countably compact whenever $$E \subseteq \bigcup_{i=1}^{\infty}G_i$$ there exists finite $$n_1...n_k$$ such that $$E \subseteq \bigcup^k_{i=1}G_{n_k}$$.

I think the following is true?

$$E$$ compact $$\iff$$ $$E$$ has Heine-Borel property

$$E$$ countably compact $$\implies$$ $$E$$ compact (Does the reverse hold?)

But are the definitions of Heine Borel and Countable Compactness really saying the same thing? It occured to be that Heine-Borel could be a stronger statement than countable compactness because the cover for $$E$$ in Heine-Borel could be a uncountable union of open sets.

Could anyone clarify my doubts?

• How do you do define compact then? Mar 19 '20 at 13:19
• My definition of compact is every sequence of points in metric space $M$ has a convergent subsequence converging to a point in $M$ Mar 19 '20 at 13:21