I am having trouble showing that $y(t)=(2\cos(2t), \sin(2t))$ is a periodic solution of the system:
$$\frac{dx}{dt}=-4y+x\left(1-\left(\frac{x^2}{4}\right)-y^2\right)$$ and $$\frac{dy}{dt}=x+y\left(1-\left(\frac{x^2}{4}\right)-y^2\right)$$
I substite $x=r\cos$ and $y = r\sin$ but I can't get the terms to cancel in order to get the solution
You just need to substitute in the equation and after some work you will succeed. Hint: $$ x'=-4\sin(2t)\quad\text{and}\quad -4y+\left(1-\frac{x^2}4-y^2\right)=-4\sin(2t)+2\cos(2t)\cdot0=-4\sin(2t). $$
But something quite interesting, which includes an answer to your question as a very particular case, is the following remark. Notice that $$ xx'+4yy'=(x^2+4y^2)\left(1-\frac{x^2}4-y^2\right). $$ For points in the ellipse $$ \frac{x^2}4+y^2=1 $$ (which in fact are the points in your periodic solution), you have $$ \frac{d}{dt}\left(\frac{x^2}4+y^2\right)=\frac12(xx'+4yy')=0, $$ which implies that you have indeed a periodic orbit. On the other hand, the sign of $xx'+4yy'$ is positive inside the ellipse and negative outside, indicating that all other solutions, other than $(0,0)$, tend to the ellipse when the time approaches $+\infty$.