Let $K$ be a nonempty closed, convex subset of $R^d$. Prove that if $x\notin K$, then there exists a unique point $y\in K$ that is closest to $x$.
My attempt: Suppose y and z are 2 different point that are both closest to x. Then,
$||x-y||=\mbox{inf}\{||x-a||: a \in S\}$ and $||x-z||=\mbox{inf}\{||x-b||: b \in S\}$. Consider the midpoint $w$ of the line that joins y to z. Then
$w=\frac{||y-z||}{2}=\frac{||y-x+x-z||}{2}\le \frac{||x-y||+||x-z||}{2}\le \frac{||x-a||+||x-b||}{2}$ for any $a \in S$. I don't know how to go from here; I thought the midpoint would be zero since we are trying to show that these 2 points are the same point, so is the midpoint formula I used incorrect? Also can you use $a$ for both infimums or do I have to use $a$ and $b$ distinctly for y and z as I did?