Why is a discrete subset of a compact space finite? I have a question about the proof of Lemma 5.13 in John Lee's text, the proof is shown below. 

My question is about the phrase underlined in red, why is a discrete subset of the compact set finite? 
Now, the author never gives a definition for discrete subset, he only gives it for discrete topology but from the phrase underlined in blue, I can tell that his definition of discrete set is as follows, 
$S$ is a discrete subset of a topological space $X$ if for each $ s \in S$, there exist a neighborhood $U$ of $X$ such that $ U \cap S = \{s\}$.  Is this correct?
Then, if you take $X$ to be the closed interval $ [-1,1]$ with the subspace topology induced from $\mathbb{R}$, then $X$ is compact, and let $S = \{ {1 \over n}, n \in \mathbb{N} \} $, then according to the definition above, $S$ is discrete but $S$ is not finite??
I feel I'm missing something but I don't know what it is.
Thank you.  
 A: You're right -- this was a mistake in the statement of Theorem 5.13. There are some corrections for this in my online errata list, which you will probably want to download and have handy when you read the book.
A: $\{\frac{1}{n}: n =1,2, \ldots\}$ is discrete in $[0,1]$ (meaning that it has the discrete topology as its subspace topology), but not finite.
If discrete is defined (as is sometimes done) as $A' = \emptyset$ (i.e. a set without limit points), where $A'$ is the derived set
$$A' = \{x \in X: \forall O \text{ open in } X: x \in O, (O\setminus \{x\}) \cap A \neq \emptyset \}\text{,}$$
then it does hold: such an $A$ is closed (as $A' \subseteq A$), so $A$ is compact when $X$ is. The set $A$ has the discrete topology as a subspace : $x \in A$ then $x \notin A'$ so there is an open set $O$ with $(O\setminus \{x\}) \cap A = \emptyset$, and this implies that $O \cap A = \{x\}$ is open in $A$. The cover $\{\{x\}: x \in A\}$ is an open cover (by relatively open subsets) of $A$, and this must be a finite cover, as we cannot omit any set in it or we wouldn't cover $A$, so $A$ is finite.
