Seems to be false. For example, let $X=Z=\mathbb R, Y=[0,1]$ and let $f$ be the zero map. Then for any $y\in [0,1]$ we have that $g(y,0)\notin \mathcal H(\mathbb R)$ so $g$ is not even defined. If you object to this example, then letting $f(x,y)=xy$ and $X=[0,1]$ we get that $g(0,0)=[0,1]$ but $g(y,z)=\{z/y\}$ for all other $y\in [0,1]$, so this is clearly not continuous.

I do not think you need a reference for this, since $|min(A)-min(B)|$ and $|max(A)-max(B)|$ are clearly upper bounds for the Hausdorff distance between $A$ and $B$.

Seems to be false. For example, let $X=Z=\mathbb R, Y=[0,1]$ and let $f$ be the zero map. Then for any $y\in [0,1]$ we have that $g(y,0)\notin \mathcal H(\mathbb R)$ so $g$ is not even defined. If you object to this example, then letting $f(x,y)=xy$ and $X=[0,1]$ we get that $g(0,0)=[0,1]$ but $g(y,z)=\{z/y\}$ for all other $y\in [0,1]$, so this is clearly not continuous.

I do not think you need a reference for this, since $|min(A)-min(B)|$ and $|max(A)-max(B)|$ are clearly lower bounds for the Hausdorff distance between $A$ and $B$.