Sequential Compactness in the space $\mathbb R^n$. I was reading the proof of the following theorem in "Analysis for Applied Mathematics" by Ward Chenney.
Then, I started thinking that if $S$ is any subset of the ball in Lemma $1$ above, then $S$ must be compact because any sequence $\{s_k\}$ in $S$ would obey the same inequality $-c\leq s_k(i)\leq c,$ so the above argument holds.
This thought then led me to think that:
 If $A$ is any bounded subset of $\mathbb R^n$ with respect to the supremum norm (as in Lemma $1$ above), then A is compact.
Am I right ?
 A: You are technically incorrect, it is because any closed subset of a compact set is compact in a metric space, i.e $(\mathbb{R}^n,\|\cdot\|_\infty)$, you have just thought ouf it this way since a finite closed ball is bounded.
Also any bounded set cannot hold, since open balls are bounded, and they are open and not compact.
What you are looking for is the "Heine-Borel" theorem, which is that any closed and bounded subset of $\mathbb{R}^n$ is compact.
A: This depends on your definition of "compact set". Usually one requires not only that sequences have limit points, but also that those limit points belong to the set. In the case of the ball above, this is assured by the fact that each component of the limit point $x^\star$ will satisfy the inequality $-c\le x^\star(i)\le c$. 
If you replace the ball with an arbitrary bounded set, then by the procedure described in your book you can prove that any sequence in that set has a limit point, but you cannot say that the limit point belongs to the set. Indeed, that's false: consider as an example the open ball $\{x\in \mathbb{R}^n\ :\ -c<x(i)<c\}$ and the sequence 
$$x_n=\left( c-\frac{1}{n}, 0, \ldots 0\right).$$
