Clarification of sequential compactness theorem with example 
A set of real numbers $S$ is said to be sequentially compact provided that every sequence $\{a_n\}$ in $S$ has a subsequence that converges to a point that belongs in $S$.

For clarification, does the definition of sequentially compact mean that for all possible sequences $\{a_n\}$ in $S$,  there exists some subsequence $\{a_{nk}\}$  that converges to some number in $S$?
To get to my main question, taken from the chapter 2.4 of the same advanced calculus textbook is an example that defines the set of real numbers $$S \equiv (0, 2]$$ and then says 

$S$ is not sequentially compact. Indeed, $\{1/n\}$ is a sequence in $S$. This sequence converges to $0$. 

Let me make sure I understand this and please correct me if I'm wrong. To determine that $S$ is not sequentially compact, we look for any possible counterexample sequence $\{a_n\}$. We take the limit $\lim_{n\to \infty}\{a_n\}$, and if we can find a subsequence that converges to a value not in $S$, then $S$ is not sequentially compact. Any subsequence would start at an index greater than $n_0$ and so any subsequence will converge to $0$.  In this example,  $\lim_{n\to \infty}\{a_n\} = 0$ for any subsequence $\{a_{nk}\}$, for $k > 0$. Since $0 \notin S$, $S$ is not sequentially compact.
 A: The answer to the first question is yes. Your argument for the main question is actually correct, since you showed that no subsequence converges to a point of $S$, but it’s not true that finding one ‘bad’ subsequence is enough to show that $S$ is not sequentially compact. You must show that every subsequence fails to converge to a point of $S$.
I’d also phrase the main argument to emphasize when I’m talking about convergence in $\Bbb R$ and when I’m talking about convergence in $S$, but that’s partly a matter of taste. The argument hinges on three facts:


*

*If a sequence $\sigma$ converges to some $x$ in some space, then every subsequence of $\sigma$ also converges to $x$ in that space. 

*No sequence in $\Bbb R$ converges to two different points.

*If a sequence converges in $S$ to some point $x$, then it also converges to $x$ in $\Bbb R$.
Here $\sigma=\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$, which converges to $0$ in $\Bbb R$, so every subsequence of it also converges to $0$ in $\Bbb R$. If $S$ were sequenctially compact, $\sigma$ would have to have a subsequence converging to some $x\in S$, but then in $\Bbb R$ that subsequence would converge both to $0$ and to $x\ne 0$, which is impossible.
Added: Note that there are other ways for a set $S\subseteq\Bbb R$ to fail to be sequentially compact. For example, $S=\Bbb Z$ fails because the sequence $\langle n:n\in\Bbb N\rangle$ in $S$ has no convergent subsequence even in $\Bbb R$, so it certainly has none in $S$.
A: You don't have it quite right. To show that the set is not sequentially compact we need to exhibit a sequence $(a_n)$ such that no subsequence of $(a_n)$ converges to a point in $S$. The sequence $(\frac{1}{n})$ does have that property, since in fact all subsequences converge to $0$, which is not in $S$.
