$\delta(x)$ is even 1-Lip (for any $f$). Indeed, if $|x-y|=a$, then $\delta(y)\geqslant \delta(x)-a$, since $(y-c,y+c)\subset (x-\delta(x),x+\delta(x))$ for $c=\max(\delta(x)-a,0)$. Analogously $\delta(x)\geqslant \delta(y)-a$ and therefore $|\delta(x)-\delta(y)|\leqslant a$.

$\delta(x)$ (defined as the maximum of appropriate $\delta$'s) is even 1-Lip (for any $f$). Indeed, if $|x-y|=a$, then $\delta(y)\geqslant \delta(x)-a$, since $(y-c,y+c)\subset (x-\delta(x),x+\delta(x))$ for $c=\max(\delta(x)-a,0)$. Analogously $\delta(x)\geqslant \delta(y)-a$ and therefore $|\delta(x)-\delta(y)|\leqslant a$.