if a sequence has a convergent subsequence, is it a Cauchy sequence in the norm? Suppose {$g_n$} has a convergent subsequence {$g_{n^k}$},how can we prove that {$g_n$} is a Cauchy sequence in the norm,I tried using triangle inequality but failed,the question is from stein's real analysis(chapter 4,lemma4.1)
 A: As I already mentioned in the comments, it is in general not correct, that if a sequence has a convergent subsequence, that it is convergent itself. Just take $(-1)^k$.
Furthermore this is not what the proof asserts. The author merely says, that if we would have a convergent subsequence of $g_n$, then we would already have a closest point $g_0$. Let me explain how this works:
Let $g_{n_k}$ be a convergent subsequence of $g_n$ and let us call the limit $g_0$. Since $\|f-g_n\|\rightarrow d$, this is also correct for all subsequences, hence
$$\|f-g_{n_k}\|\rightarrow d$$
On the other hand, the convergence of $g_{n_k}$ w.r.t. to the norm yields
$$\|f - g_{n_k}\|\rightarrow \|f-g_0\|.$$
Hence we have
$$\|f-g_0\| = d.$$
Since $S$ is closed, we also have $g_0\in S$. Hence $g_0$ is a closest point and the result would follow.
These kind of arguments can be quite powerful. For example it did not need, that $S$ is a subspace, just that it is closed. In other words, if you can somehow find such a convergent subsequence, you can find closest points also for more general $S$.
