Proof regarding the periodicity of a continuous-time sinusoid after sampling
You might be making this problem harder than it is by thinking in terms of frequency rather than periods.  Your sinusoid is
$$x(t) = \cos\left(2\pi \frac{t}{T_p}\right)$$ and its samples are
$$x[n] = x(nT) = \cos\left(2\pi \frac{nT}{T_p}\right).$$
The samples form a discrete-time periodic sequence of period $N$ if 
\begin{align}
x[n] &amp;= x[n+N]~\forall\, n\\
&amp;{\Large \Downarrow}\\
\cos\left(2\pi \frac{nT}{T_p}\right) &amp;= \cos\left(2\pi \frac{(n+N)T}{T_p}\right)\\
&amp;{\Large \Downarrow}\\
\cos\left(2\pi \frac{nT}{T_p}\right)&amp;= \cos\left(2\pi \frac{nT}{T_p} + 2\pi \frac{NT}{T_p}\right)
\end{align}
where the last equality will hold only if $\displaystyle\frac{NT}{T_p}$ equals some integer $k$. This is equivalent to the OP's desired requirement that $$\frac{T}{T_p} = \frac kN$$ for some integers $k$ and $N$.

So, what is the fundamental period of $x[n]$? That is,
what is the smallest integer $N_0$ such that $\displaystyle\frac{N_0T}{T_p}$ is an integer?  Well, suppose that the ratio $\displaystyle\frac{T}{T_p}$ is the rational number $\frac{k_1}{k_2}$ where $k_1$ and $k_2$ are relatively prime integers. Then, $N_0 = k_2$ and $\displaystyle\frac{N_0T}{T_p} = k_1$.  Thus, considering $x[n]$ as a periodic discrete-time sequence of samples spaced $T$ seconds apart  -- think of it as the impulse train $\sum_n x(nT)\delta(t-nT)$ -- gives a (continuous-time) period of $k_2T$ seconds for the sequence/impulse train. Note that one period of the discrete-time sequence spans $k_1$ periods of the sinusoid of period $T_p$.