Although this is a standard derivation, you frequently don't see it in introductory electromagnetism courses, maybe because those courses shy away from the heavy use of vector calculus. Here's the usual approach. We'll find a wave equation from Maxwell's equations.
Start with
$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$.
Take a partial derivative of both sides with respect to time. The curl operator has no partial with respect to time, so this becomes
$\nabla \times \frac{\partial\vec{E}}{\partial t} = -\frac{\partial^2\vec{B}}{\partial t^2}$.
There's another of Maxwell's equations that tells us about $\partial\vec{E}/\partial t$.
$\nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$
Solve this for $\partial\vec{E}/\partial t$ and plug into the previous expression to get
$\nabla \times \frac{(\nabla \times \vec{B})}{\mu_0\epsilon_0} = -\frac{\partial^2 \vec{B}}{\partial t^2}$
the curl of curl identity lets us rewrite this as
$\frac{1}{\mu_0 \epsilon_0}\left(\nabla(\nabla \cdot \vec{B}) - \nabla^2\vec{B}\right) = -\frac{\partial^2 \vec{B}}{\partial t^2}$
But the divergence of the magnetic field is zero, so kill that term, and rearrange to
$\frac{-1}{\mu_0 \epsilon_0}\nabla^2\vec{B} + \frac{\partial^2 \vec{B}}{\partial t^2} = 0$
This is the wave equation we're seeking. One solution is
$\vec{B} = B_0 e^{i (\vec{x}\cdot\vec{k} - \omega t) }$.
This represents a plane wave traveling in the direction of the vector $\vec{k}$ with frequency $\omega$ and phase velocity $v = \omega/|\vec{k}|$. In order to be a solution, this equation needs to have
$\frac{\omega^2}{k^2} = \frac{1}{\mu_0\epsilon_0}$.
Or, setting $v = 1/\sqrt{\mu_0\epsilon_0}$
$\frac{\omega}{k} = v$
This is called the dispersion relation. The speed that electromagnetic signals travel is given by the group velocity
$\frac{d\omega}{d k} = v$
So electromagnetic signals in a vacuum travel at speed $c = 1/\sqrt{\mu_0\epsilon_0}$.
Edit
You can follow the same steps to derive the wave equation for $\vec{E}$, but you will have to assume you're in free space, i.e. $\rho = 0$.
Edit
The curl of the curl identity was wrong, there's a negative number in there
Author gives a clue on the transition:
Let us assume that $\delta\vec{A}$ vanishes at infinity and integrate (formula (1)) by parts...
This is the usual step in the Lagrangian theory of field (actually, of anything). At first, we have the variation of action written in an awkward form:
$$\delta S=\int_{\substack{\text{domain of least}\\\text{action problem}}}(\text{something})\cdot(\text{derivative of }(\text{variation}))$$
Since we want the variation of action to be in form $\delta S=(\text{something})\cdot(\text{variation})$, we have to "pull out" the variation from under the derivative. This is done by integration by parts:
$$\delta S=\Bigl[(\text{something})\cdot(\text{variation})\Bigr]_{\text{boundary of domain}}\\-\int_{\text{domain}}(\text{variation})\cdot(\text{derivative of }(\text{that something}))$$
And then we use the fact that the variation is put to be zero on the boundary of the domain (in this case, at infinity). That means that the first term cancels out, and we have finally
$$\delta S=-\int_{\text{domain}}(\text{variation})\cdot(\text{derivative of }(\text{something}))$$
Exactly what we need to proceed to $(\text{derivative of }(\text{something}))=0$.
A side note: I found a typo in the book, comparing the Russian and English editions. In the English edition the formula (1.6) is typesetted as
$$\delta S=\int \mathrm{d}t\,\mathrm{d}^3\delta x\mathbf{A}(t,\mathbf{x})\mathbf{F}[\mathbf{A}_0(t,\mathbf{x})]+\mathcal{O}(\delta A^2)$$
which for me hardly makes any sense (what is the differential of the variation, and what is $\delta x$ in the field case in the first place?). Actually, in the Russian edition this formula looks like
$$\delta S=\int dt\,d^3x\,\delta\mathbf{A}(t,\mathbf{x})\mathbf{F}[\mathbf{A}_0(t,\mathbf{x})]+O(\delta A^2)$$
which is rather more comprehensible. No wonder you stumbled over this. My condolences.
Best Answer
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.