How to derive Maxwell's equations from the electromagnetic Lagrangian? In Heaviside-Lorentz units the Maxwell's equations are:
$$\nabla \cdot \vec{E} = \rho $$ 
$$ \nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J}$$
$$ \nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 $$
$$ \nabla \cdot \vec{B} = 0$$
From EM Lagrangian density: $$\mathcal{L} = \frac{-1}{4} F^{\mu \nu}F_{\mu\nu} - J^\mu A_\mu$$ 
I can derive the first two equations from the variation of the action integral: $S[A] = \int \mathcal{L} \, d^4x$. Is it possible to derive the last two equations from it?
 A: From a geometric perspective, the last two equations are a consequence of:
(1)F = dA  (Faraday tensor is the exterior derivative of the four-potential)
(2) dd = 0 (the exterior derivative of the exterior derivative vanishes)
Thus
(3) dF = 0
which gives the last two equations in your question.
A: Assume for simplicity that the speed of light $c=1$. The existence of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ alone implies that the source-free Maxwell equations
$$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"}$$
$$ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$$ 
are already identically satisfied. To prove them, just use the definition of the electric field 
$$\vec{E}~:=~-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t},$$
and the magnetic field 
$$\vec{B}~:=~\vec{\nabla}\times\vec{A}$$ 
in terms of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$.   
The above is more naturally discussed in a manifestly Lorentz-covariant notation.
OP might also find this Phys.SE post interesting. 
Thus, to repeat, even before starting varying the Maxwell action $S[A]$, the fact that the action $S[A]$ is formulated in terms the gauge $4$-potential $A^{\mu}$ means that the source-free Maxwell equations are identically satisfied. Phrased differently, since the source-free Maxwell equations are manifestly implemented from the very beginning in this approach, varying the Maxwell action $S[A]$ will not affect the status of the source-free Maxwell equations whatsoever.
A: Note that since $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, we have the equation
$$
\partial_\mu F_{\nu\alpha} + \partial_\alpha F_{\mu\nu} + \partial_\nu F_{\alpha\mu} = 0
$$
This  equation is called the Bianchi Identity. This equation is separate from the equations of motion one obtains from varying the action. It can be shown that the Bianchi equation is equivalent to the last two equations you have mentioned in your question.
NOTE: To elaborate on @Qmechanic's answer, the very fact that $F_{\mu\nu}$ can be written as $\partial_\mu A_\nu - \partial_\nu A_\mu$ is itself a consequence of the last two equations.                        
