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 $B$