Is possible reach the DFT if I have the DTFT? My teacher told me that DFT is DTFT sampled, i.e.: 

$$X[k] = X(e^{j \omega})\Bigg|_{\omega = \frac{2\pi k}{N}}$$ 

But, if I have the sine

$$ x[n] = \sin(\omega_0 n) $$

the DTFT is: 

$$X(e^{j \omega}) = \frac{\pi}{j} \big(\delta(\omega - \omega_0) - \delta(\omega + \omega_0)\big)$$ 

(periodic in $\omega$ with period $2\pi$). How is it possible to reconcile this with the DFT for $N$ values for this sine or any signal?

Thanks people. Sorry for the English.
The DFT is a sampled version of the DTFT only for finite length signals. Otherwise, there is no point in comparing the DTFT with the DFT because you can only compute the DFT for finite length (or periodic) signals. Your example $x[n]=\sin(\omega_0n)$ is an infinitely long sequence. For specific choices of $\omega_0$ it is periodic, but nevertheless, you can't obtain its DFT by sampling its DTFT. The operation of sampling a Dirac impulse is undefined because $\delta(\omega)$ cannot be evaluated at $\omega=0$ since the Dirac impulse is not an ordinary function but a distribution.