Building on Beni's answer, suppose that this is not right-continuous, i.e. we have some $\delta$ such that $\sup\{|f(x)-f(y)| : d(x,y)\leq \delta+\epsilon, f\in \mathcal{F}\}-\sup\{|f(x)-f(y)| : d(x,y)\leq \delta, f\in \mathcal{F}\}>z$ for some fixed $z>0$ and for arbitrarily small $\epsilon>0$. Then for any fixed $\epsilon$ we have some f\in \mathcal{F},x',y'\in \mathbb{R} such that d(x',y')\leq \delta+\epsilon and |f_\epsilon(x')-f_\epsilon(y')|-\sup\{|f_\epsilon(x)-f_\epsilon(y)| : d(x,y)\leq \delta\}>z meaning that if we let x'',y''\in\mathbb{R} be such that d(x'',y'')\leq \delta, d(x',x'')\leq \epsilon,$d(y',y'')\leq \epsilon$ (which can always be done) we get |f_\epsilon(x'')-f_\epsilon(x')| + |f_\epsilon(y'')-f_\epsilon(y')| \geq |f_\epsilon(x')-f_\epsilon(y')| - |f_\epsilon(x'')-f_\epsilon(y'')|>z so one of the summands on the left must be at least $z/2$, meaning that for sufficiently small $\epsilon>0$ (as in smaller than $z/2$) and any $\delta>0$ we have some function $f_\delta\in\mathcal{F}$ for which $d(x,y)<\delta\not\Rightarrow d(f_\delta(x),f_\delta(y))<\epsilon$ and so $\mathcal{F}$ is not equicontinuous.