Bolzano-Weierstrass property in a normed vector space I am trying to prove the following:
Let $(V,||.||)$ be a normed vector space. Show that V has the property that any bounded sequence in V has a convergent subsequence if and only if $S=\{v\in V: ||v||=1\}$ is compact.
This is my attempt so far:
For '$\Rightarrow$': Suppose $U$ is an open cover for $S$. Take a bounded sequence $(x_n) \subseteq V$. Then $(y_n)=(x_n)/||x_n||$ is in $S$ and $(y_n)$  has a convergent subsequence $(y_{n_r})$. I would then like to construct a finite subcover of $S$ using $(y_{n_r})$- perhaps using the fact that it is a Cauchy sequence? But I don't think this will get me anywhere...
For '$\Leftarrow$': I make use of the following proposition 'If a sequence in a metric space has no convergent subsequences, then for each  $x\in X$ there exists $\epsilon_x >0$ such that $B(x,\epsilon_x)$ contains $(x_n)$ for finitely many n.'
Take $(x_n) \in V$, where $(x_n)$ is bounded and consider, $(y_n)=(x_n)/||x_n|| \subseteq S$. Suppose, for a contradiction, that $(y_n)$ has no convergent subsequence. Then $\forall v \in S 
, \exists \epsilon_v$ such that $B(v,\epsilon_v)$ contains $(y_n)$ for finitely many n. Clearly $\{ B(v,\epsilon_v): v \in S\}$ is an open cover for S, so there is a finite subcover. But each open ball in this finite subcover contains $(y_n)$ for finitely many n, so S contains $(y_n)$ for finitely many n, which contradicts the fact that $(y_n)$ is a sequence in S. Hence $(y_n)$ has a convergent subsequence $(y_{n_r})$ in S. Then $(x_{n_r}) =(y_{n_r}) \times ||(x_{n_r})||$ also converges, since $||(x_{n_r})||$ is bounded- I'm not really sure about this last step.
Any hints/help would be much appreciated!
 A: Hint: In a normed space compactness and sequential compactness coincide. Your proof for the converse is correct, once you make the last step precise. What do you know about the product of a convergent with a bounded sequence?
Edit (Additional explanation for the first hint): A subset $K \subseteq X$ of a topological space $X$ is called sequentially compact if every sequence contained in $K$ has a convergent subsequence. If the topology (i.e. the open sets) of $X$ comes from a metric, a subset $K \subseteq X$ is compact iff it is sequentially compact. You should have seen this statement for $\mathbb{K}^n$ in your course on real analysis. Convince yourself that the proof still works for an arbitrary normed space. Proving sequential compactness of the sphere $S$ is then not too hard with your assumption.
Sidenote: in a general topological space the notions of sequential compactness and compactness are entirely unrelated. Ultimately the reason for this is that sequences can only probe for countable properties and spaces can get too big for this. One then considers nets instead, which then gives equivalence of net compactness and compactness also in the general case.
