Assume $(M,g)$ is Schwarzschild manifold, $A$ is the minimum area in the homology class of the boundary. How to prove that $A=$ the area of the horizon.

I'm interpreting your question in the only reasonable manner that I can think of:
This clearly is a stronger than the same question in "exterior" Schwarzschild, i.e. $M\backslash B_{m/2}(0)$ with the same metric there. The answer is yes. I'll give a sketch because I'm short on time: Let $w=1+\frac{m}{2r}$. $\Delta_\delta w = 0$ by computation (it has to hold if you look up the formula for conformal change of scalar curvature and know that Schwarzschild is scalar flat). Then, if $\Sigma$ is a "competitor" which is homologous to $S_{m/2}$, an appropriate Green's theorem shows that $$ \int_{S_{m/2}} D_\nu w = \int_\Sigma D_\nu w $$ These integrals are all with respect to the $\delta$quantities. Here is where I will skip the details. Multiply by some constant $C$ (which you should explicitly determine) so that the LHS becomes $area(S_{m/2},g_m)$. That is $$ area(S_{m/2}(0),g_m) = C\int_{S_{m/2}} D_\nu w = C\int_\Sigma D_\nu w $$ Now, I'll leave it up to you to check that $CD_\nu w$ is bounded from above by $w^4$ (you just need to compute the various quantities). Thus, we've shown that $$ area(S_{m/2}(0),g_m) \leq \int_\Sigma w^4 = area(\Sigma,g_m). $$ You may further check that equality implies that $\Sigma = S_{m/2}(0)$. This is a classic example of a "calibration" argument. I'll remark that Hugh Bray, in his thesis gave a beautiful argument to show that the other centered spheres are isoperimetric. These ideas have had a good deal of influence in related works, e.g. EichmairMetzger proved a quantitative isoperimetric inequality in Schwarzschild and used this to study large isoperimetric surfaces in initial data which is $C^0$asymptotic to Schwarzschild. Also, BrendleEichmair used this, along with other results to study isoperimetric surfaces in the "doubled" Schwarzschild, as I defined above. 

