Baby Rudin Theorem 2.27c I feel like I'm missing something very simple here, but I'm confused at how Rudin proved Theorem 2.27 c:
If $X$ is a metric space and $E\subset X$, then $\overline{E}\subset F$ for every closed set $F\subset X$ such that $E\subset F$. Note: $\overline{E}$ denotes the closure of $E$; in other words, $\overline{E} = E \cup E'$, where $E'$ is the set of limit points of $E$.
Proof: If $F$ is closed and $F \supset E$, then $F\supset F'$, hence $F\supset E'$. Thus $F \supset \overline{E}$.
What I'm confused about is how we know $F \supset E'$ from the previous facts?
 A: If $A \subset B$ then $A' \subset B'$.
Pf:  If $a \in A'$ then $a$ is a limit point of $A$.  So every neighborhood of $a$ contains a point $b \in A$ with $b \ne a$.  But if $b \in A$ then $b \in B$ as $A \subset B$.  So every neighborhood of $a$ a conains a point $b \in B$ with $b \ne a$.  So $a$ is a limit point of $B$.  And $a \in B'$. 
And $A' \subset B'$.
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So if $F \supset F'$, and $F\supset E$.  Then $F' \supset E'$ and $F \supset F' \supset E'$.
A: If $x$ is a limit point of $E$ then $x = \lim x_n$ for some sequence $x_n \in E \setminus \{x\}$. If $E \subseteq F$ then $x_n \in F \setminus \{x\}$ so we can also say that $x$ is a limit point of $F$. Therefore
$$ E' \subseteq F' \subseteq F. $$
A: $F\supset F'$ because  $F $ is closed.   $F'\supset E'$ because  $F\supset E $, by assumption. Therefore $F\supset E' $.
A: It's not too hard to prove that for any sets $A \subseteq B$ in a metric space $(X, d)$, it follows that $A' \subseteq B'$. With the above result at hand, we can prove Theorem $2.27$(c).
Proof: To prove this theorem choose a closed set $F$ that contains $E$, this means that $E \subseteq F$. 
Observe that since $F$ is closed $F = \overline{F} = F \cup F'$. Also note that since $E \subseteq F$ it follows that $E' \subseteq F'$. It then follows by elementary set theory that $E \cup E' \subseteq F \cup F'$. But this implies that $\overline{E} \subseteq \overline{F} = F$ (since $\overline{E} = E \cup E'$) which concludes the proof. $\square$

As an aside, Rudin's exposition can be very terse at times, so don't be discouraged if some things seem simple at first but you can't quite flesh out their details.
A: Let $x \in E'$, then $\forall r > 0$, $N_r(x) \cap(E-\{x\}) \ne \emptyset$. Since $E \subset F$, then $\forall r > 0$, $N_r(x) \cap(F-\{x\}) \ne \emptyset$.Thus, $x \in F'$. But F is closed. So, $F' \subset F$. So, $x \in F$. Thus, $E' \subset F$. Finally, $\bar{E} \subset F$.
A: Let's go by contradiction. We know from assumption that $E \subset F$, thus we just need to prove the limit points of $E$ are also in $F$, i.e. $E' \subset F. $ Assume $x\in  E'$ but $x\notin F$. Since $x\notin F$ therefore $x\in F^c$ and $F$ is closed, therefore $F^c$ is open. This means there is a neighborhood of $x$, $N_r(x)$ that is contained only in $F^c$ and does not intersect with $F$ (i.e. $N_r(x) \cap F = \emptyset$). This is a contradiction due to $E \subset F$, since according to definition of limit point, every neighborhood of $x$ intersects with $E$.
