Is Gibbs energy change during osmosis positive? If so then why is it spontaneous? For ideal solution, μA=μA* +RT
  lnxA


ln function is an increasing function, means that the higher the xA the higher the μA......(1)

Now consider osmosis. From Atkins' Physical Chemistry Ninth Edition, page 173,


  osmotic pressure, π, is the pressure that must be applied to the solution
  to stop the influx of solvent (into solution across a semi-permeable
  membrane).


Here is my interpretation and thought process:

Had you not exerted π on solution, the μ(solution)&lt; μ(solvent) based on (1) because lnxA is negative for xA&lt;1, so by Second Law, the osmosis is spontaneous and solvent should move from pure solvent (high μ) to solution (low μ). Sounds right. 

However, during this process, μ(pure solvent) unchanged (μ is intensive, not affected by amount) but μ(solution) increases due to dilution. Based on (1), for the solution, since xA increases μ(solution) must also increase. Although the process stops when μ(solution)=μ(pure solvent), the effect is that μ increases. μ increases would in turn imply ΔG>0 which means non-spontaneous, but as far as I know osmosis is spontaneous, so where is my misconception?
Let the subscript A refer to the solvent and B refer to the solute, and let the subscript 1 refer to the pure solvent container and  2 refer to the mixture container.  Then the free energy of the combined system is given by:
$$G=n_{A1}\mu_A^0+n_{A2}\left(\mu_A^0+RT\ln{\left[\frac{n_{A2}}{n_{A2}+n_{B_2}}\right]}\right)+n_{B2}\left(\mu_B^0+RT\ln{\left[\frac{n_{B2}}{n_{A2}+n_{B_2}}\right]}\right)$$If the number of moles of solvent in container 2 increases by $dn_{A2}$ and the number of moles of solvent in container 1 decreases by $dn_{A2}$, the change in the free energy of the combined system changes by dG.  If we do the math correctly, we obtain:
$$dG=RT\ln{x_{A2}dn_{A2}}\tag{1}$$where $x_{A2}$ is the mole fraction of solvent in the mixture container at any point:  $$x_{A2}=\frac{n_{A2}}{n_{A2}+n_{B_2}}$$Eqn. 1 indicates that the change in free energy is negative (as expected).