Distance between a cone and a disjoint hyperplane I seek to prove the following, which I guess is true:

Define $A:=\{x \ge 0\} \subset \mathbb{R}^m$ and assume that $U\subset \mathbb{R}^m$ is an affine subspace with $A \cap U=\emptyset$. Show that
$$ \delta:=\inf_{\substack{x\ge 0\\x'\in U}}\lVert x-x'\rVert>0.$$

I can easily handle the cases where $\mbox{dim}(U)=0$ or $1$.
 A: I'll assume $U$ is non-empty. $U$ is of the form $u+V$ for some point $u$ and some linear subspace $V.$ Consider the projection map $\phi:\mathbb R^m\to \mathbb R^m/V.$ It takes $U$ to a single point which I will denote $[u].$  The distance between $A$ and $U$ is bounded below by (in fact equal to) the distance between $\phi(A)$ and $[u].$ So it remains to show that the distance between $\phi(A)$ and $[u]$ is positive.
Each standard basis vector $e_i$ is taken to a point other than $[u]$ because $e_i\in A$ which is disjoint from $U.$ Since $A$ is a convex cone generated by the rays $e_1,\dots,e_m,$ the set $\phi(A)$ is a convex cone generated by the rays $\phi(e_1),\dots,\phi(e_m).$ A finitely generated convex cone is polyhedral (the "Farkas-Minkowski-Weyl theorem", see for example Schrijver's "Theory of Linear and Integer Porgramming") and therefore closed. The distance between a point $[u]$ and the closed set $\phi(A)$ disjoint from $[u]$ is positive, which is what we needed to prove.
A: Not sure how to solve this,  but I can help you with one more case. If U is a hyperplane then there is some vector y and constant c such that $U=\{ x: x \cdot y =c \}$. Disjointness implies that $y \geq 0, c<0$. This implies the distance is at least $ - c/ \| y \| $. 
