I'm having a little trouble with the following exercise:
Let $V \subset\mathbb{R}^p$ be a non-empty, closed set and $a \in \mathbb{R}^p$. For $x, y \in \mathbb{R}^p$ we note $d(x,y) = \|x-y\|$. Furthermore, let $d(a,V) = \inf \{\, d(a,x) \mid x \in V \,\}$. Show that there exists $b \in V$ such that $d(a,V) = d(a,b)$. Hint: consider the set $V \cap \overline{B(a; R)}$ for a suitably chosen $R > 0$.
Now I know that $a \in V \Leftrightarrow d(a,V) = 0$, so if $a \in V$ we can simply choose $b = a$. However, I am not quite sure how to prove this statement for $a \notin V$, and I don't really know how to use the hint that was given. I already looked at similar questions, but they all include theorems about compact spaces, and in this exercise $V$ isn't necessarily compact (or so I think). Any hints would be very much appreciated.
The claim is false unless we add the facepalm condition that $V$ is not empty. So assuming $V\ne 0$, we can pick $v\in V$, let $R=\|a-v\|$, and as suggested consider $W:=V\cap \overline{ B(a;R)}$. This set is closed and bounded, hence compact. Therefore, the continuous function $W\to \Bbb R$, $x\mapsto d(a,x)$ attains its minimum at some point $b\in W$. Then $d(a,b)=d(a,W)$. Verify that this also equals $d(a,V)$.