Help with Metric spaces $\newcommand{\Int}{\operatorname{Int}}\newcommand{\Bdy}{\operatorname{Bdy}}$
If $A$ and $B$ are sets in a metric space, show that:
(note that $\Int$ stands for interior of the set)


*

*$\Int (A) \cup \Int (B) \subset \Int (A \cup B)$.

*$(\overline{ A \cup B}) = (\overline A \cup \overline B )$.  (note that $\overline A = \Int (A) \cup \Bdy(A)$ )


Now for the first (1) I see why its true for instance in $R$ we can have the intervals set $A=[a,b]$ and $B=[b,c]$ we have $A \cup B=[a,c]$ so $\Int(A \cup B)=(a,c)$ now $\Int(A)=(a,b)$ and 
$\Int(B)=(b,c)$ so we lose $b$ when we take union to form $\Int(A) \cup \Int(B)=(a,b) \cup (b,c)$.
 A: For number two, I would recommend demonstrating it with Venn Diagrams.  Have two overlapping circles labeled A and B.  Via shading, show that the same space is shaded for each side of the equation.
A: $\newcommand{\Int}{\operatorname{Int}}\newcommand{\Bdy}{\operatorname{Bdy}}$You have made a good start on (1). However, given two intervals $[a, b]$ and $[c, d]$ on $\mathbb{R}$ why should it be the case that $b = c$? Moreover, why do two sets on $\mathbb{R}$ even need to be closed intervals?
What if $A = \mathbb{N}$. Then we still have $A \subset \mathbb{R}$!
Thus you should go back to the definition of the interior! 
Suppose $x \in \Int(A) \cup \Int(B)$. What does this mean? Once we know what it means, can we show why $x \in \Int(A \cup B)$? In this way, you will also avoid needing any of the special characteristics of $\mathbb{R}$!
A: Recall that $\operatorname{Int}(A) \subset A$ for any set $A$ and that the interior is the largest such open set contained in $A$.  Similarly, $\overline A$ is the smallest closed set containing $A$.
Then for $(a)$, $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subseteq A \cup B$ and is an open set, so $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subseteq \operatorname{Int}(A \cup B)$.
For $(b)$, $\overline{A} \cup \overline{B}$ is a closed set containing $A \cup B$, so $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$.  For the other side of the inclusion, since $\overline{A} \subseteq \overline{A \cup B}$ and $\overline{B} \subseteq \overline{A \cup B}$, we have $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$.
