Is possible reach the DFT if I have the DTFT?
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.