Bolzano-Weierstrass on arbitrary finite dimensional vector spaces Let $V$ be a finite dimensional real, normed vector space, of dimension $n$. Now, we know that $V$is isomorphic to $\mathbb{R}^{n}$ and so if we have a sequence $\{x_{n}\}$ in $V$ such that $||x_{n}|| \leq C$ for some real number $C > 0$, then we know that $||x_{n}||$ is a bounded sequence of real numbers, and so (by BWT) has a convergent subsequence. But my question is that: doesn't the convergent subsequence of real numbers converge not to a vector $x \in V$, but only to the value of a norm $||x||$ (since we're taking the norms as a sequence of real numbers), meaning that any vector with that norm could be our limit? How can we use Bolzano-Weierstrass to argue for convergence to a specific vector in $V$?
 A: Your argument is wrong. Even though the sequence $\Vert x_n \Vert$ has a limit, this does not yet imply that the sequence $x_n$ has a limit.
(This can be seen by considering the sequence $e_1, e_2, e_3, \dots$ in the infinite-dimensional space $\ell^1$: Your assumption is true (the norms converge to one), but the conclusion is not: the sequence diverges. Of course, this is not a counterexample to the Bolzano-Weierstrass theorem because $\ell^1$ is infinite-dimensional - but it shows that your conclusion is too fast.
Actually, if we want to prove the Bolzano Weierstrass theorem in higher (but finite) dimensions, we need to be more clever than just consider the norm. E.g., you could use the fact that $\Vert x_n \Vert$ is bounded to show that each component (after choosing a basis, of course) of each vector $x_n$ is bounded. Thus, you can iteratively construct a subsequence such that the first components converge, which has a subsequence such that the first and second components converge, which has a subsequence ... you get the idea.
After finding such a convergent subsequence, we can then move on to prove that this particular subsequence $x_{n_k}$ satisfies $ \lim_{k \rightarrow \infty}\Vert x_{n_k}\Vert$, not the other way around as you suggested.
