How does Bolzano Weierstrass imply every closed and bounded set is sequentially compact Bolzano Weierstrass states that:

A. Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence

Why does it imply that (from Difference between Heine-Borel Theorem and Bolzano-Weierstrass Theorem) 

B. Every closed and bounded set is sequentially compact

Problem 1. The word set does not appear in A, but appear in B. Is every set a sequence and every sequence a set? I find the $\Rightarrow$ part unintuitive, but it is clear that every sequence is a set.
Problem 2. The word closed doesn't appear in A, but appears in B.  If a bounded sequence has a convergent subsequence, then the "set" associated with the sequence is closed. Does this follow from the completeness of $\mathbb{R}^n$?
So my question is just what is the relationship between a bounded sequence and a closed bounded set. 
 A: Resolution 1. The fact that set does not appear in (A) is fine. Certainly not every set is a sequence. Sequential compactness of a set $A$ means that every sequence in $A$ has a subsequence which converges in $A$. If $A$ is bounded, that means that every sequence therein is a bounded sequence. Bolzano Weierstrass gives us that each sequence has a subsequence which converges in $\mathbb R^n$. But if we also assume that $A$ is closed, that subsequence converges in $A$. Therefore, $A$ is sequentially compact. The key here is that we have a set, and we have to prove things about sequences in that set. That is how the sequences from Bolzano Weierstrass help with the sets in sequential compactness.
Resolution 2. If a bounded sequence has a convergent subsequence, it does not mean that the sequence must be closed. Consider, for instance, the sequence $\{1/n\}$. Every subsequence converges to $0$, but the sequence is not closed as a set. But this is still okay. Bolzano Weierstrass says that a bounded sequence has a subsequence which converges in $\mathbb R^n$. If that sequence lives in a non closed set, then the point to which the subsequence converges may not be in that set. But since a closed set contains all of its limit points, any convergent subsequence in a closed set converges in the set. We need this because a set being sequentially compact requires that the subsequence converges inside the set.
