What you wrote is not entirely correct. What you should've used is the reverse triangle inequality $$|\lVert x_k\rVert-\lVert a\rVert |\leq \lVert x_k-a\rVert\leq M$$
This gives that $$-M\leq\lVert x_k\rVert-\lVert a\rVert\leq M$$ $$\lVert a\rVert-M\leq\lVert x_k\rVert\leq M+\lVert a\rVert$$
so the sequence is bounded. It seems strange to use Bolzano Weiertrass, since what you're being asked to prove is precisely that theorem. I would argue as follows. You probably know the result of Bolzano Weiertrass for $\Bbb R^1$. But one can extend it easily to $\Bbb R^n$. Consider a bounded sequence in $\Bbb R^n$ $${\bf x}_k=(x_{k,1},x_{k,2},\ldots,x_{k,n})$$
By Bolzano Weiertrass, $x_{k,1}$ has convergent subsequence, $y_{k,1}=x_{n_k,1}$. Now look at $${\bf x}_{n_k}={\bf y}_k=(y_{k,1},y_{k,2},\ldots,y_{k,n})$$
This is now a subsequence of the whole ${\bf x}_k$. We know the subsequence $y_{k,2}=x_{n_k,2}$ is bounded, so it has a convergent subsequence by Bolzano Weiertrass in $\Bbb R^1$, which we will call $z_{k,2}=y_{k_j,2}$. Now we have a subsequence of ${\bf y}_k$ (which was already a subsequence of ${\bf x}$, whose two first coordinates have matching subindices and both converge. And as you should see by now, we ultimately obtain a subsequence of "deepness $n$" (meaning it will be the subsequence of a subsequence of ... a subsequence of ${\bf x}$ - $n$ times) which we can call $${\bf x'}_j=(x_{j,1}^\prime,\ldots, x_{j,n}^\prime)$$ which converges, since each coordinate converges. Thus, Bolzano Weiertrass is proven for $\Bbb R^n$. Since closed balls are closed sets, your claim follows, since closed sets contain their limit points.
NOTE Bolzano Weiertrass is precisely the statement that every closed ball in $\Bbb R^n$ is compact, equivalently sequentially compact, or that $\Bbb R^n$ is locally compact, that is, every point $a\in\Bbb R^n$ has a neighborhood whose closure is compact.
Best Answer
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.