Is my answer correct? I'm trying to solve this question:

My solution:
Since $\varphi$ is continuous we have:
$C\text{ is convex}\implies C\text{ is connected}\implies \varphi(C)\text{ is connected}\implies \varphi(C) \text{ is an interval}$.
Am I right? I didn't use the fact the norm is induced by the inner product and in the chapter I'm studying the author haven't written yet about continuous function.
I found also the question a little bit weird, a set can have more than one infimum?
Thanks 
 A: You've shown that the function has at most one inf (and achieves it if the domain is compact, although you didn't mention that part). The problem is to show that there's at most one point in the domain at which the function takes on that value. 
Consider the function $f: [0, 1] \to [0, 1] : x \mapsto 1$.
That function has a minimum  (namely "1"), but it takes on that min value at every point of the domain. You have to show that YOUR function isn't like this. 
In short: your proof does not prove the thing you were asked to prove. 
A: My first answer addressed the question you asked ("Is my proof correct?"). This one suggests how you might proceed with a correct proof:
Suppose that there were two points, $a, b$ whose squared distance to $p$ was minimal. Consider the function 
$$
h(t) = \phi((1-t)a + tb) - \phi(a).
$$
Then $h(0) = h(1) = 0$. Can you evaluate $h(0.5)$, for instance? What can you say about the points $(1-t)a + tb$, for $t \in [0, 1]$? 
Further hint: the algebra will be simplified a bit if you just assume that $p$ is the origin; for the non-origin case, first translate everything by $-p$ so that $p$ becomes the origin. Translation leaves distances invariant, so you've reduced the problem to a simpler case. 
