How is a continuous spectrum for the DTFT possible?
The reason the DTFT has values at frequencies that result in discrete aperiodic sinusoids is because we are interested in representing all possible discrete time signals including (linear combinations of) discrete aperiodic sinusoids. Let's say that we have a signal $x[n] = e^{j \omega_0 n}$ where $\omega_0$ is not a rational multiple of $\pi$. Would you like this signal to have a DTFT? Of course. It would not have one if we did not allow the existence of the DTFT at all frequencies (including irrational multiples of $\pi$). 

Put another way, periodicity is not a requirement for the existence of the DTFT of a signal.