A Question on Bolzano-Weierstrass Theorem

Question

The following are two equivalent statements about the Bolzano-Weierstrass theorem.

1. Every bounded real sequence has a convergent subsequence.
2. A subset of $\mathbb R$ is compact if and only if it is closed and bounded.

Now consider the set $A :=\mathbb Q \cap [0, 1]$, since rationals are countable, we can treat $A$ as a bounded sequence from $0$ to $1$. Then we get the following conclusion for each statement listed above.

1. There is a convergent subsequence in $A$. For example, $a_n := 1-\frac{1}{n}, \ n \in \mathbb N$.
2. $A$ is not compact since it is not closed.

My question is where closedness in the second statement plays a role in the first statement. In other words, we know for every sequence from a compact set has a convergent subsequence. If we treat a real sequence as a set, called $B$, then statement $1$ says there is a convergent subsequence in $B$ so long as $B$ is bounded. Again, the closedness condition is missing. Why? Thank you!

Answer 1

Bolzano-Weierstrass theorem says that a set $K\subset \mathbb R$ is compact if every sequence has a convergent subsequence in $K$ (i.e. the limit of the convergent sequence must be in $K$), so $K$ is closed.

Answer 2

I'm not sure what the question here is, but (1) requires that any sequence must have a convergent subsequence, and in $\;A\;$ we can find a sequence which does not converge to a point in $\;A\;$ , for example

$$\left(1-\frac1n\right)^n\xrightarrow[n\to\infty]{}\frac1e\notin A\;,\;\;yet\;\;\forall\,n\in\Bbb N\;,\;\;\left(1-\frac1n\right)^n\in A$$

Of course, we must know that if a real sequence converges then any subsequence converges and to the same point.