Short answer
You've got half of the answer right when mentioning Faraday's Law of Induction, which tells us how electric fields can be generated from time-varying magnetic fields. The other half of the answer involves Ampère's Law with Maxwell's correction, which tells us how magnetic fields can be generated either from an electric current or time-varying electric fields. This coupled interaction between the electric and magnetic fields allows the EM waves to propagate.
Long answer
The previously stated claim is typically proven mathematically in most standard textbooks written on classical electromagnetism, and so such derivations can readily be found in the appropriate literature. I will give one such derivation here, adapted from the Wikipedia article that you can and should read in order to learn more about EM radiation in fuller detail:
Consider Maxwell's Equations in the microscopic form (in SI units),
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \tag{1}$$
$$\nabla \cdot \mathbf{B} = 0 \tag{2}$$
$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \tag{3}$$
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \tag{4}$$
with (3) and (4) being, respectively, the Maxwell-Faraday Equation and the Maxwell-Ampère equation.
In free space (i.e. in a location that contains no electrical charges or currents), the equations take the form,
$$\nabla \cdot \mathbf{E} = 0 \tag{5}$$
$$\nabla \cdot \mathbf{B} = 0 \tag{6}$$
$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \tag{7}$$
$$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \tag{8}$$
Taking the curl of (7), we end up with,
$$\begin{align} \nabla \times \left(\nabla \times \mathbf{E}\right) & = \nabla \times \left(-\frac{\partial \mathbf{B}}{\partial t}\right) \\
& = - \frac{\partial}{\partial t} \left(\nabla \times \mathbf{B}\right) \\
& = - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} \end{align} \tag{9}$$
where we have made a substitution in the last step using (8).
From vector calculus, for a vector field $\mathbf{F}$, $\nabla \times \left(\nabla \times \mathbf{F}\right) = \nabla \left(\nabla \cdot \mathbf{F} \right) - \nabla^2 \mathbf{F}$, where $\nabla^2$ is the vector Laplacian operator, is an identity. Using this identity to rewrite (9),
$$\nabla \left(\nabla \cdot \mathbf{E} \right) - \nabla^2 \mathbf{E} = - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} \tag{10}$$
But since we're in free space, by equation (5) the first term on the left-hand side of (10) vanishes, and after redefining $\mu_0 \epsilon_0 = 1/c^2$ we're left with,
$$\nabla^2 \mathbf{E} - \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 \tag{11}$$
which is clearly the wave equation with the electric field $\mathbf{E}$ as the dependent function.
Taking the curl of (8) and performing an almost identical procedure, the wave equation for the magnetic field $\mathbf{B}$ can also be derived,
$$\nabla^2 \mathbf{B} - \frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} = 0 \tag{12}$$
The fact that Maxwell's Equations in free space directly lead to these wave equations shows that the coupling between electric and magnetic fields described by Faraday and Ampère directly allows for the forward propagation of electromagnetic waves.