Let $\langle\cdot , \cdot \rangle$ be an inner product in $\mathbb{R}^m$ and $U \subset \mathbb{R}^m$ such that $U$ is compact, convex and $0 \notin U$. Show that we can find a vector $v \in \mathbb{R}^m$ and a number $\delta > 0 $ such that $\delta \leq \langle u , v \rangle$ for all $u \in U$.
My attempt: Since $0 \notin U$ then $0 \in \mathbb{R}^m - U$. Since $U$ is compact, it is closed, and then $\mathbb{R}^m - U$ is open. Then there is $\delta > 0$ such that the open ball $B(0, \delta) \subset \mathbb{R}^m - U$. But then we can find a vector $v \in \mathbb{R}^m$ such that $v \in B(0, \delta)$ but $v \notin U$. And I do not know how to proceed from here.
Any help would be great.
Hint : since $U$ is compact, it admits a point $p$ closest to $0$. Consider $v=p$. Is it possible to find $q \in U$ with $\langle q,p \rangle < \langle p, p \rangle =: \delta$ ? (Think of a small perturbation $p + \epsilon (q-p)$)