Proving $ \overline{E \cup F} = \overline {E} \cup \overline{F}$ and $ \overline{E \cap F} \subset \overline {E} \cap \overline{F}$ Let $E, F \subset X$, prove that $ \overline{E \cup F} = \overline {E} \cup \overline{F}$. 
For further clarification: I'm referring to $\overline{E}$ as E closure. E' would be the limit point, defined as $E' = \{ ( E \cap N_r (p) ) \backslash \{p\} \neq \emptyset , \ \  \forall R >0 \}$. This is also in a topological space. 
I proved it as follows:

$\overline{E \cup F} = \overline{E \cup E' \cup F \cup F'}$
  $\\ = \overline{(E \cup E') \cup (F \cup F')}$
  $\\ = \overline{(E \cup E')} \cup \overline{(F \cup F')}$
  $\\ = \overline{(E)} \cup \overline{F}$

The next one is a little tricky for me. Prove $ \overline{E \cap F} \subset \overline {E} \cap \overline{F}$. Here's my go at it:

Let $x \in \overline{E \cap F}$. Then $x \in \overline{ (E \cup E') \cap (F \cup F)} \rightarrow x \in \overline{E} \cap \overline{F}. $

Feedback would be much appreciated! I've also been trying to come up with examples that would help me visualize these statements a little better. 

To prove that $\overline{E \cap F} \subset \overline{E} \cap \overline{F}$:
Let $x \in \overline{E \cap F}$. There exists $(x_n)_n \subset E \cap F$ such that $x_n \underset{n \rightarrow \infty}{\rightarrow} x$.
$(x_n)_n \subset E \cap F$ means that $(x_n)_n \subset E$ and also that $(x_n)_n \subset F$. Can you finish this?
