Show that $\sup_y \inf_x h(x,y) \leq \inf_x \sup_y h(x,y)$ 
Here is what I have so far:
Since $h(x,y)$ has a bounded range, $\exists \sup, \inf$ for range $h(x,y)$. Then $h(x,y) \leq \sup \{h(x,y): y \in Y\} = f(x)$ and $h(x,y) \geq \inf\{h(x,y) : x \ in X \} = g(y)$. Thus we get the inequality $$g(y) \leq h(x,y) \leq f(x) \forall x \in X, \forall y \in Y $$
Now we see that $f(x)$ is an upper bound of $g(y)$. Here, I am not sure what I am trying to show from here. I think the last line reads that we want to show that the $\sup$ with respect to y * $\inf h(x,y) \leq$ $\inf$ with respect to x * $\sup h(x,y)$. Is this at least the right direction for thinking of the problem?

Your proof of the inequality $g(y) \leq f(x)$ is correct. Fix $x\in X$. As $g(y) \leq f(x)$ for all $y\in Y$, $f(x)$ is an upper bound of $\{ g(y) : y\in Y\}$. It follows that
$$
\sup \{ g(y) : y\in Y\} \leq f(x).
$$
As the above inequality holds for arbitrary $x\in X$, $\sup \{ g(y) : y\in Y\}$ is a lower bound of $\{ f(x) : x\in X\}$ and we get
$$
\sup \{ g(y) : y\in Y\} \leq \inf \{ f(x) : x\in X\}.
$$
