Suppose $S$ is a subset of $\mathbb R$ and $x \in \mathbb R$. Show that there are at most two nearest points of $S$ to $x$
Attempt:
Suppose there are three nearest points $a,b,c$ of $S$ to $x$. Let $dist(x,S) =t= \inf \{~~ d(x,s) ~|~s\in S \} $ represent the distance of $x$ from $S$.
Now, by the given assumption : $d(a,x)=d(b,x)=d(c,x)=t=dist(x,S)$
Strategy $1$: If we bring a contradiction that when there are $3$ or more nearest points to $S$, one of them, say, $d(a,x)>t$, then, we can prove that there can be atmost $2$ nearest points.
Strategy $2$: If we prove that $a=b$ and $b \ne c$ by proving that $d(a,b)=0,$ then also, we can prove it.
The second way of going seems somehow better to me.
I know this result $dist(x,S) \le |x - \sup S|$ , with equality when $x > \sup S$
Could anyone please tell me how to proceed ahead.
Thank you very much for your help in this regard.