For $u \in W^{1, \infty}(\mathbb R^n)$, $|u(x) - u(y)| \le \|\nabla u \|_\infty |x - y|$

Question

Let $u \in W^{1, \infty}(\mathbb R^n)$, $\varphi \in C^\infty_0(\mathbb R^n)$ and $$\varphi_\varepsilon(x) = \frac{\varphi(x/\varepsilon)}{\varepsilon^n}.$$ We define $u^\varepsilon = \varphi_\varepsilon \star u \in C^\infty(\mathbb R^n)$. By the mean value theorem, we deduce that $$|u^\varepsilon(x) - u^\varepsilon(y)| \le \|\nabla u^\varepsilon\|_\infty|x - y|.$$ I have shown, using the fact that $\nabla(\varphi \star u) = \varphi \star \nabla u$ and Holder's inequality, that $$\|\nabla u^\varepsilon\|_\infty \le \|\nabla u\|_\infty,$$ so that $$|u^\varepsilon(x) - u^\varepsilon(y)| \le \|\nabla u\|_\infty|x - y|.$$ Now I don't see how I can show that $$|u(x) - u(y)|\le |u^\varepsilon(x) - u^\varepsilon(y)|.$$ Is it the right way to do that ? Any hint ?