As explained by Pacciu, |u(x)|\leqslant\|u'\|_\infty\,|x-y|+|u(y)| for every $x$ and $y$. Assume first that $\varepsilon\leqslant\frac14$ and $x\geqslant2\varepsilon$ and let us integrate this inequality from $y=x-2\varepsilon$ to $y=x$. The result is 2\varepsilon |u(x)|\leqslant\|u'\|_\infty\,\int_{x-2\varepsilon}^{x}|x-y|\mathrm dy+\int_{x-2\varepsilon}^{x}|u(y)|\mathrm dy\leqslant\|u'\|_\infty\,2\varepsilon^2+\|u\|_1. The same inequality holds if $x\leqslant2\varepsilon$, using the integral from $y=x$ to $y=x+2\varepsilon$.
Thus, if $\varepsilon\leqslant\frac14$, \|u\|_{\infty}\leqslant\varepsilon \|u'\|_\infty+\frac1{2\varepsilon}\|u\|_1. In particular, \|u\|_{\infty}\leqslant\tfrac14\|u'\|_\infty+2\|u\|_1 hence, for every $\varepsilon\geqslant\frac14$, \|u\|_{\infty}\leqslant\varepsilon \|u'\|_\infty+2\|u\|_1.
Finally, for every $\varepsilon\gt0$, \|u\|_{\infty}\leqslant\varepsilon\|u'\|_\infty+\max\left\{2,\frac1{2\varepsilon}\right\}\|u\|_1.