Nyquist Rate of cosine modulated function
Update: I feel there is no need to bring concept of Bandpass sampling (as mentioned in comments below) because this is specifically regarding a simple misunderstanding that OP has. The text referenced asks about Nyquist Rate (which wrongly mentions it as Nyquist Frequency) but OP asks about highest Frequency in $y(t)$.

For the component $\frac{1}{2}X(\Omega-\Omega_0)$ the highest frequency content is at $\Omega_0+\Omega_N/2$. The lowest (starting frequency) is at $\Omega_0-\Omega_N/2$. Similary for the other component $\frac{1}{2}X(\Omega+\Omega_0)$, the starting frequency is at $-\Omega_0-\Omega_N/2$ and ending frequency is at $-\Omega_0+\Omega_N/2$. 

So if you see the overall frequency content of $y(t)$, lowest = $-\Omega_0-\Omega_N/2$ and highest = $\Omega_0+\Omega_N/2$. This is your signal $y(t)$ whose highest frequency is $\Omega_0+\Omega_N/2$.

The Nyquist rate for this is $2\Omega_0+\Omega_N$ if you consider $y(t)$ as baseband signal in the picture below. But the minimum sampling rate will be much lesser than this if you consider it as Passband signal (concept of bandpass sampling as mentioned in another answer)


  If the Highest frequency of $X(\Omega)$ is $\Omega_N$


No. The highest frequency content in this case is $\Omega_N/2$ which is half the Nyquist rate. Otherwise this would cause aliasing.