Zero-order interpolation problem
You're thinking too complicated. 

For a digital signal to be periodic, it simply has to repeat every $P\in\mathbb N$ samples.

Now, you have a continuous signal which is $\frac{2\pi}{\omega_0}$-periodic. 

Obviously, if a period happens to be exactly "worth" a natural number $N\in\mathbb N$ sample times, the resulting sample from sampling it will repeat with a period of $N$. 

More generally, if $M\in\mathbb N$ periods of the continuous signal are as long as $N$ sampling times, the digital signal will also be periodic.

So, as a general rule, one period of the input signal must be worth $\frac{N}{M}$ sampling times, or:

$$\begin{align*}
\frac{2\pi}{\omega_0} &amp;\overset != \frac NM \frac {1}{f_\text{sampl}}&amp;\quad N,M\in \mathbb N\\
&amp;= R\frac {1}{f_\text{sampl}} &amp;\quad R\in \mathbb Q\text,\\
\text{which means that}\\
\frac{f_\text{sample}}{\omega_0} &amp;= \frac {R}{2\pi}\text;\\
\text{with}\\
f_{sample}&amp;=\frac{\omega_s}{2\pi}\\
\text{it follows that}\\
\frac{\omega_s}{\omega_0} &amp;= R\text,
\end{align*}$$

i.e. the relationship between signal frequency and sampling rate needs to be rational.

The fact that you have a zero-order interpolator afterwards doesn't change this consideration -- the interpolator's output will be periodic iff its input is.