First, \begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . ..\end{align} Subtract: $$\mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt$$ So it suffices to show that this integrand is positive. It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.