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.
Jasper’s answer contains the key idea, but it also contains a fair bit of unnecessary material. Start with your sequence $\big\langle\langle x_n,y_n\rangle:n\in\Bbb N\big\rangle$ in $X\times Y$. You know that $\langle x_n:n\in\Bbb N\rangle$ has a convergent subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ in $X$, say with limit $x$. Now consider the sequence $\langle y_{n_k}:k\in\Bbb N\rangle$ in $Y$: it has a convergent subsequence $\langle y_{n_{k_j}}:j\in\Bbb N\rangle$ in $Y$, say with limit $y$. Now show that $$\Big\langle\left\langle x_{n_{k_j}},y_{n_{k_j}}\right\rangle:j\in\Bbb N\Big\rangle$$ converges to $\langle x,y\rangle$ in $X\times Y$.
Best Answer
Take the sequence {$ 1,2,3,...$} in $\mathbb R$ . Does it have any convergent subsequence?