It's false. Take $M = [0,1]$ and $N = \mathbb{R}$ and define $F(x,y) = 1 - |x-y|$. Taking $y_x = x$ satisfies condition (2). Here $\inf_M \sup_N F(x,y) = 1$ and $\sup_N \inf_M F(x,y) = 1/2$, achieved when $y = 1/2$.

It's false. Take $M = [0,1]$ and $N = \mathbb{R}$ and define $F(x,y) = 1 - |x-y|$. Taking $y_x = x$ satisfies condition (2). Here $\inf_M \sup_N F(x,y) = 1$ and $\sup_N \inf_M F(x,y) = 1/2$, achieved when $y = 1/2$.

Edit: this answers the original question. The new version of the question, with $N$ compact, is falsified by taking $N=[0,1]$ in the above example.