This is a textbook proof. From this textbook: http://a.co/d/dC9s4px
Assume that every open covering of $S$ has a finite subcovering. Let $\{ a_j \}$ be a sequence in $S$. Assume, seeking a contradiction, that the sequence has no subsequence that converges to an element in $S$. This must mean that for every $s \in S$ there is an $\epsilon_s > 0$ such that no element of the sequence satisfies $0 < |a_j - s| < \epsilon_s$. Let $I_s = (s - \epsilon_s, s + \epsilon_s)$. The collection $\mathcal{C} = \{ I_s \}$ is an open covering of the set $S$. By hypothesis, there exists a finite subcovering $I_{s_1}, \ldots, I_{s_k}$ of open intervals that cover $S$. But then $S \subseteq \cup_{j=1}^k I_{s_j}$ contains no element of the sequence $\{ a_j \}$, and that is a contradiction.
As the textbook says "for every $s \in S$ there is an $\epsilon_s > 0$ such that no element of the sequence satisfies $0 < |a_j - s| < \epsilon_s$", but $s$ itself may or may not be in the sequence $\{ a_j \}$. Then each interval $I_s = (s - \epsilon_s, s + \epsilon_s)$ will have zero or one unique sequence elements from $\{ a_j \}$. And the finite subcovering of $I_{s_1}, \ldots, I_{s_k}$ can have up to $k$ unique elements from the sequence $\{ a_j \}$, not zero as the textbook says.
A sequence with a finite number of unique elements must have a subsequence that converges to one of those elements, which provides the contradiction to complete the proof. However, the textbook still seems wrong to claim that the finite subcovering will cover zero elements of the sequence.
Am I correct on this or not?
You did not explicitly mention that you assume $S \subset \mathbb{R}$ nor did you specify the definition of compactness. There are various definitions of compactness for topological spaces, and it seems that you understand a topological space $X$ to be compact if every sequence in $X$ has a convergent subsequence. In modern language we say that $X$ is sequentially compact. Historically this concept was introduced for subsets $X$ of an Euclidean space $\mathbb{R}^n$. It turned out, however, that the concept of compactness in the realm of general topological spaces is much better defined via covers: A space $X$ is called compact if each open covering of $X$ has a finite subcover.
The proof in your textbook is not correct. We have to proceed as follows.
Let $(a_j)$ be a sequence in $S$. Assume it has no convergent subsequence. Then for each $s \in S$ there exists $\epsilon_s > 0$ such that $\lvert a_j -s \rvert < \epsilon_s$ only for finitely many $j$. W.l.o.g. we could assume that $0 < \lvert a_j -s \rvert < \epsilon_s$ for no $j$, but is still possible that $a_j = s$ for finitely many $j$. The collection $\{ I_s \}_{s \in S}$ is an open cover of $S$. By hypothesis, there exist finitely many $s_k$ such that $S = \bigcup_{k=1}^k I_{s_k}$. But each $I_{s_k}$ contains only finitely many $a_j$, therefore also $\bigcup_{k=1}^k I_{s_k} = S$ contains only finitely many $a_j$. This is a contradiction.
You can easily generalize this proof to show that if a metric space $X$ is compact, then it is sequentially compact.
Note that for metric spaces also the converse is true.