Show that if $A\subseteq B$, then inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$ Is my logic correct or am I missing something?
Show that if $A\subseteq B$, then inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$
Case 1. If $A \subset B$ then there exists $b_1,b_2\in B$ such that $b_1,b_2 \notin  A$. Let $a_1,a_2 \in A$ such that $a_1=$ inf A and $a_2$ = sup A. Suppose $b_1$=inf B and $b_2$=sup B than $b_1<a_1<a_2<b_2$ which implies inf $B<$ inf $A < $ sup $A<$ sup $B$
Case 2. If $A = B$ than every element in $A$ is in $B$. This implies that if $A$ is bounded above or below so is $B$ and vice versa. If the sup $B$ is defined to be the least upper bound and the inf $B$ is defined to be the greatest lowest bound. Than sup $B$ = sup $A$ and inf $B$ = inf $A$. 
Since $A \subseteq B$ the following equality can be written as inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$
 A: Let's proceed by contradiction.
1)  inf B $\le$ inf A
Assume, to the contrary, that inf B > inf A.  Then there is some $x$ such that inf A < $x$ < inf B, but then $x \in A $ and $x \notin B$, a contradiction since $A \subseteq B$.  (Also note here: If I were being super rigorous I would need to appeal to the definition of infimum to make sure that $x$ is indeed in $A$.  Notice the strict inequality.)
Thus,  inf B $\le$ inf A
2)  sup A $\le$ sup B
Assume, to the contrary that sup A > sup B, then there exists some $x$ such that sup B < $x$ < sup A.  However, then $x \in A$ and $x \notin B$.  Again, a contradiction for the same reason as before.
3)  inf A $\le$ sup A.  By definition of sup and inf.
All together inf B $\le$ inf A $\le$ sup A $\le$ sup B.
Maybe someone can post a comment to help a little bit, what property of the real numbers am I using when I assume existence of $x$ like I do above?  Certainly it must exist since there are no gaps in the real numbers.  That way I can say that given two real numbers $a,b$ with $a <b$ then there exists $x$ such that $a <x <b$.  Is this the Archimedian property?  something topological?
A: Your logic is incorrect in Case 1: You've assumed that $b_1 = \inf{B} \in B \setminus A$. Take $A = (0, 1)$ and $B = (0, 2)$ to see why this is quite problematic.

For a hint towards the correct direction, can you show that any lower bound of $B$ is a lower bound of $A$ as well? So the greatest lower bound of $B$ is still a lower bound for $A$?
A: let's suppose A ⊆ B⊆R  bounded  then we can say that through the known theorem for real numbers Completeness Property of R:
for every a ∈ A : infA ≤ a ≤ supA(1)
likewise for every b ∈ B : infB ≤ b ≤ supB
but at the same time A is the subset of B so we can say that every element of A is also bounded by the least upper bound  and the greatest lower bound of B
for every a ∈ A : infB ≤ a ≤ supB
but from the definition of the supremum and infimum we know that for random upper bound x and lower bound  y the supremum and infimum will always be smaller and greater likewise
for every a ∈ A : supA ≤x      and    y≤ infA (2)
in our problem we can assume the random bounds of x an y as x=supB   , y=infB
Through (1) and (2) we have our solution
inf B ≤ inf A ≤ sup A ≤ sup B
