Let $\mathbb{R}^d$ be a the usual Euclidean space and let $Y$ be a fixed non-empty closed subset of $Ball(0,1)$ (the unit ball in $\mathbb{R}^d$ about $0$ of radius $1$). 


Let $f$ be the map taking $x \in \mathbb{R}^d$ to the hyperplane passing through $0$ and $x$.  

Is the map $g$ defined by
$$
g(x)\triangleq f(x)\cap Y,
$$
continuous from $\mathbb{R}^d$ to the Hausdorff space $H(Ball(0,1))$? 


If not, can we at-least conclude that it is measurable with respect to an atomless measure on $H(Ball(0,1))$?



Intersection with a fixed set in Hausdorff metric space
Here is what is true: Let $h(p)$ be the hyperplane through $0$ with normal vector $p$ for each $p\neq0$. Then the function $p\mapsto H(p)=h(p)\cap B\_1(0)$ is continuous. 


Indeed, $H\_p=\{x\in\mathbb{R}^d\mid px=0,\|x\|\leq 1\}$ and the Hausdorff distance of $H\_p$ and $H\_q$ is bounded by $\max\_{x\in B\_1(0)}|(p-q)x|$. Since the latter is a function continuous in $(p-q)$ and vanishes at $0$, the result follows. 


Every sensible continuous selection that makes the original problem well-defined should lead to the same outcome. 


