Bridging CTFT and DTFT for a cosine
You're almost there. First, there's a small scaling error in the transform $X_c(\omega)$. Since the Fourier transform of an impulse train is given by

$$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\omega_s\sum_{k=-\infty}^{\infty}\delta(\omega-k\omega_s)\tag{1}$$

with $\omega_s=2\pi/T$, and since multiplication in the time domain corresponds to convolution in the frequency domain with a scaling of $1/2\pi$ (if you use $\omega$ as the independent frequency variable), the transform of the sampled signal is

$$X_c(\omega) = \frac{\omega_s}{2} \sum_{n=-\infty}^{\infty} \big[  \delta(\omega - \omega_x - n\omega_s) +  \delta(\omega + \omega_x - n\omega_s) \big]\tag{2}$$

Now you just need to use the relation between $\omega$ and $\Omega$

$$\omega=\frac{\omega_s}{2\pi}\Omega\tag{3}$$

and the scaling property of the Dirac Delta impulse

$$\delta(ax)=\frac{1}{|a|}\delta(x)\tag{4}$$

which for your example results in

$$\delta(\omega)=\frac{2\pi}{\omega_s}\delta(\Omega)\tag{5}$$

and you'll arrive at the correct expression for the DTFT of a sampled cosine.

Also take a look at this answer to a related question.