Convex sets. Real Analysis. 
Let $C \subset \mathbb{R^{n}}$ a convex set. Fixed $p \in \mathbb{R}^{n}$, let $\varphi: C \longrightarrow \mathbb{R}$ the function defined by $\varphi(x) = |x-p|=\sqrt{\langle x-p,x-p\rangle}$. Theres exists as most one point $a \in C$ such that $\varphi(a) = \inf \lbrace \varphi(x) | x \in C\rbrace$.

I don't know how to relate convexity to other hypothesis. I thought suppose $a \neq b$ such that $\varphi(a) = \varphi(b) = \inf\lbrace \varphi(x) | x \in C\rbrace$ and use $\varphi(a) \leq \varphi(b)$, $\varphi(b) \leq \varphi(a)$ so $\varphi(a) = \varphi(b)$. But I can't do anything with it anymore. I appreciate any hint.

You need C to be closed btw. Notice that $\phi(x)$ is strictly convex. So if there were two minimizers then there would be a point $z$ between them with strictly lower value $\phi(z)$. Also, there must be a minimum on C if C is closed.
