 Solution

For $x\ne y$, $|({\delta }_x-{\delta }_y)(f)|\le 2\Vert f{\Vert }_{\infty }$, giving an upper bound $2$ for the operator norm. By Urysohn lemma, there is $h:K\to [0,1]$ with $h(x)=1,h(y)=0$. Then $f=2h-1$ has uniform norm $1$ and $({\delta }_x-{\delta }_y)(f)=2$. Thus every distinct pair has operator norm $2$, andThe stated hypotheses also permit a singleton $K$. Then there are no distinct pairs, so the infimum is $+\infty$ under the extended-real convention, or undefined as a real number. This edge case cannot be assigned the value $2$.