$S\subseteq T$ means every member of $S$ is a member of $T$.
$a = \inf S$ means $a$ is the largest lower bound of $S$, so it's $\le$ every member of $S$ and $\ge$ all other lower bounds of $S$.
If a number $b$ is $\le$ every member of $T$, and every member of $S$ is a member of $T$, then $b$ is $\le$ every member of $S$. Therefore $\inf T$ is $\le$ every member of $S$. Therefore $\inf T$ is a lower bound of $S$. Therefore $\inf T$ is $\le$ the largest lower bound of $S$. In other words $\inf T\le \inf S$.
A similar argument shows $\sup T\ge\sup S$.
The statement that $\inf S\le \sup S$ is true only if $S\ne\varnothing$. If $S\ne\varnothing$, then there exists $s\in S$. And we must then have $\inf S\le s\le\sup S$.
The fact that if $S\subseteq T$ then $\inf S\ge\inf T$ shows that $\inf\varnothing\ge$ all other "inf"s, so $\inf\varnothing=\infty$. Similarly $\sup\varnothing=-\infty$.
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?
Best Answer
If $A$ and $B$ are bounded subsets of $\mathbb R$, then $\sup A\leq \inf B$ is equivalent to the statement that for all $a\in A$ and $b\in B$, $a\leq b$. Thus, it suffices to show that for each $x\in X$ and $y\in Y$, $g(y)\leq f(x)$.
Let $x_0\in X$ and $y_0\in Y$ be fixed but arbitrary. Then $g(y_0)\leq h(x_0,y_0) \leq f(x_0)$.