Let $(X,d)$ be a metric space and $A \subset X$ a subset.
Definition 1:
We call $A$ coverage compact, if each open cover of $A$ has a finite subcover.Definition 2:
We call $A$ sequentially compact, if every sequence in $A$ has a convergent subsequence.Bolzano-Weierstrass-Theorem:
$A$ is coverage compact iff $A$ is sequentially compact.
My Proof of "$\implies$":
Let $A$ be coverage compact and $(U_i)_{i \in I} \subset X$ a open cover of $A$ and $(x_k)_{k \in \mathbb{N}} \subset A$ a sequence.
Then there exists a finite subset $K \subset I$ so that $A \subset \bigcup_{i \in K} U_i := \hat{U}$.
If there would exist a subsequence of $x_k$ converging to $x^* \in A$, in each of the finitely many $U_i$ for $i \in K$ there could only be finitely many members of $x_k$ since otherwise there would be a converging subsequence.
But this is a contradiction to the fact that $x_k$ has infinitely many members.
Best Answer
No, it is not correct: