Proving principle of the Iterated Suprema

I am omitting the $x \in X$, etc, for simplicity.

$\sup_{x',y'} h(x',y') \ge h(x,y)$ for all $x,y$. Hence $\sup_{x',y'} h(x',y') \ge G(y)=\sup_{x'} h(x',y)$ for all $y$, hence $\sup_{x',y'} h(x',y') \ge \sup_{y'}G(y')$. Similarly for $F$.

Let $\epsilon>0$, then we have $x_1,y_1$ such that $h(x_1,y_1) \ge \sup_{x',y'} h(x',y') - \epsilon$. Clearly $\sup_{y'}G(y') \ge G(y_1) \ge h(x_1,y_1) \ge \sup_{x',y'} h(x',y') - \epsilon$. Since $\epsilon$ was arbitrary, we have $\sup_{y'}G(y') \ge \sup_{x',y'} h(x',y')$. Similarly for $F$.