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I have a question regarding this proof my professor gave us. For the third property, I understand the proof up to the sentence "If $x \in E'$, i.e., x is a limit point of E." Well, I also understand that if x is not in F, then x can't be a limit point since F is closed. After that, I don't fully understand it. Could someone please give me an explanation?

Thank you in advance.

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    If E subset F then any limit point of E most be a limit point of F as well as every neighborhood containing a a point of E contains a point (the same point) of F. So if x is not a limit point of F, it can't be a limit point of E. – 2017-02-22
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    I can't read sideways. Better if you rotated it or quoted from your notes. – 2017-02-22
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    @user254665 I apologize for the inconvenience. I uploaded the question from my phone and it seems StackExchange won't allow me to rotate it. – 2017-02-22

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Before we even start, notice if $E\subset F $ means $E' \subset F'$ because....

If every neighborhood of $x $ contains a point of $E $ that very same point of E is also a point of $F $ so every neighborhood of $x $ contains a point of $F $.

Now that comment you don't understand (and to tell the truth, I didn't either from the notes) doesn't matter. If $x\in E'$ either $x\in E \subset F $ or $x\in E' \subset F' \subset F $.

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    Okay, so $R \subset F$ means $E' \subset F'$ because the set of all limit points of E are contained in F. I understand why every neighborhood of x contains a point if F. Then, if $x \in E'$ we have what you stated at the end. And it we take the union of E and E', we get E-bar in F. OMG! It makes sense! – 2017-02-22
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    Thank you so much! I guess I'll just ignore what my professor said at the end. – 2017-02-22
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If $x$ is a limit point of $E$, that means there exists a sequence $(x_n)\subset E$ for which $x_n\neq x$ for all $n$, and $\lim x_n=x$. Notice that the same sequence $(x_n)\subset F$ satisfies the same conditions, so $x$ is a limit point of $F$. Since $F$ is closed, it contains its limit points. So $x\in F$.