Decimating a discrete time cosine function
The following correction of Jason R's answer completely resolves the problem: The frequency spectrum of $X_2(ω)$  consists of two components, which are scaled versions of $X(ω)$. Those are $4\pi$-periodic instead of being $2\pi$-periodic. When those $4\pi$-periodic versions are added up, the sum becomes $2π$-periodic again. THERE IS NO ALIASING or IMPULSES SUPERIMPOSED ON EACH OTHER. The derivation from
$$
X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\frac{\omega}{2}-\omega_0\right) + \delta\left(\frac{\omega}{2}+\omega_0\right) + \delta\left(\frac{\omega}{2}-\omega_0-\pi\right) + \delta\left(\frac{\omega}{2}+\omega_0-\pi\right)\right)
$$
to
$$
X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right) + \delta\left(\omega-2\omega_0-2\pi\right) + \delta\left(\omega+2\omega_0-2\pi\right)\right)
$$
is not correct. Instead it must read 
$$
X_2(ω)=π(δ(ω−2ω_0 )+δ(ω+2ω_0)+δ(ω−2ω_0−2π)+δ(ω+2ω_0−2π)) \qquad (∗) 
$$ 
This is due to the scaling property of impulses 
$$
δ(aω)=\frac{δ(ω)}{|a|}   
$$ 
Then, observing that (*) is $2π$-periodic, two of the impulses are redundant and hence the DTFT of $x_2[n]=x[2n]$  becomes
$$
X_2(ω)=π(δ(ω−2ω_0)+δ(ω+2ω_0)) 
$$