If $F \subseteq \mathbb{R}$ is closed, countable infinity. Prove that $F$ has a countless infinite number of isolated points. To prove it use the following lemma : If $F$ is closed and $x \in F$ then $x$ is a isolated point of $F$ if and only if $F-\{x\}$ is closed. 
In a book, a solution is as follows : 
Let's suppose that $\{ a_1,a_2, \cdots ,a_n \}$ is the set of all isolated points of F. Then by the lemma : $F-\{a_1,a_2, \cdots ,a_n \}$ is closed. 
Then this set which is closed and infinite countable does not have isolated points, a contradiction since everything closed countable has some isolated point.
My question comes in this last part, the statement that the set $G=F-\{a_1,a_2, \cdots ,a_n \}$ does not have any isolated point does not seem to be true.
Why? , because although it is true that the set $G$ can not contain an isolated point of $F$, that does not mean that the same set by itself has an isolated point.
I would like to know if my doubt is true or not.
My prove is identical until that last part. I affirm that there is an isolated point of $G$ let's call it $a$. So exists $\delta_1>0$ such that $(a-\delta_1,a+\delta_1) \cap G =\{a\}$. Let´s put $\delta_2 = min_{1\leq i \leq n}\{ \vert a-a_i \vert \}$. Then for $\epsilon = min\{\delta_1,\delta_2\}$ we have to $(a-\epsilon,a+\epsilon)\cap F =\{a\}$ so we would get an isolated point of $F$ different from all the others $a_i$.

Your findings agree with the proof; you have proved that if $G$ has an isolated point then $F$ has an isolated point different from all the $a_i$. But by assumption (toward a contradiction), the $a_i$ are all the isolated points of $F$. Hence $G$ does not have an isolated point.
