Let $(M,\rho)$ be a metric space, $F\subset M$ a closed subset, $\varepsilon>0$.
On page 8 of the book "Convergence of Probability measures" by Patrick Billingsley one says that the function $$f:M\to [0,1]\\ x\mapsto(1-\rho(x,F)/\varepsilon)^+$$ is uniformly continuous since it holds $|f(x)-f(y)|\leq \rho(x,y)/\varepsilon$.
Actually I'm not able to prove this inequality: I tried to solve all the possible cases ($x,y\in F, \not\in F, \in F^\varepsilon, \not\in F^\varepsilon$) but it doesn't looks elegant at all and I even could't prove all the possibilities.
$(.)^+:=max(0,.)$
Thanks