Consider the random vector $\vec v=(X,Y)$, which is a normal random vector whose covariance matrix is $\sigma^2 I$. The amazing property of the (multivariate) Gaussian distribution is that it is rotationally symmetric whenever the covariance matrix is isotropic, as it is in this case. Rotational symmetry implies that the direction that the vector $\vec v$ points in is uniformly distributed on the circle, **regardless** of what the magnitude of $\vec v$ is. More precisely, it means that the conditional distribution of the angle is independent of the magnitude, which is equivalent to saying that they are independent.

At this point, the burning question is: **why** is the distribution of $\vec v$ rotationally symmetric? One way to see it is by changing to polar coordinates. Indeed, by definition of the normal distribution, we have that for any measurable subset $A$ of the plane,
$$
\mathbb P(\vec v\in A)=\frac{1}{2\pi}\int_{A}e^{-x^2/2\sigma^2}e^{-y^2/2\sigma^2}\ dx\ dy
$$
$$
=\frac{1}{2\pi}\int_{A}e^{-r^2/2\sigma^2}\ (r\ dr\ d\theta).
$$
Here, we see that the integrand does not depend on $\theta$, which implies the rotational symmetry.

Finally, we can answer the question: $U=\|\vec v\|^2$ is a function of the **magnitude** and $V=\cos\angle \vec v$ is a function of the **angle**, so by the first paragraph these quantities are independent.