Denote by $Lip_0(X)$ the set of all Lipschitz functions on a metric space $X$ vanishing at some base point $e \in X$. The norm in $Lip_0$ is defined as fololows $$ \|f\|_{Lip_0} := Lip(f), $$ where $Lip(f)$ denotes the Lipschitz constant. With pointwise operations $f \vee g := \max\{f,g\}$ and $f \wedge g := \min\{f,g\}$ the space $Lip_0$ becomes a Lipschitz lattice, in which the following condition holds $$ \|f \vee g\|_{Lip_0} \leq \max\{\|f\|_{Lip_0},\|g\|_{Lip_0}\}. $$ The Banach lattice condition $|f| \leq |g| \implies \|f\| \leq \|g\|$, however, fails. (Nik Weaver. Lipschitz Algebras, 2nd ed.)
Question. Are operations $f_+ := f \vee 0$, $f_- := (-f) \vee 0$ and $|f| := f \vee (-f)$ continuous in the $Lip_0$ norm, i.e. does, e.g., $$ \|f_+ - g_+\|_{Lip_0} \leq C\|f - g\|_{Lip_0} $$ hold?
I have searched a lot for either a proof or a counterexample, but couldn't find anything. Any help will be appreciated.