Proving that $\boldsymbol \nabla \times (U(r) \hat{r}) = 0 $ I was just checking to see if I wsa doing this right, as it isn't a formal proof. Just showing the identity. 
Let $U(r) \hat{r}$ b a vector in spherical coordinates. Given that the vector is only dependent on r, and there's no $\hat{\phi}$ or $\hat{\theta}$ component, that means that in the usual formula for the curl in spherical coordinates: $$\nabla \times U(r)\hat{r} = \hat{r} \frac{1}{\sin \theta}\left[\frac{\partial}{\partial\theta}(u_{\phi}\sin\theta) - \frac{\partial u_{\theta}}{\partial\theta} \right]+ \hat{\theta} \frac{1}{r}\left[\frac{1}{\sin \theta}\frac{\partial u_r}{\partial\phi}- \frac{\partial}{\partial r}(ru_{\phi}) \right]+\hat{\phi} \frac{1}{r}\left[\frac{\partial}{\partial\phi}(ru_{\theta})- \frac{\partial u_r}{\partial r}\right]$$
the $u_{\theta}$ and $u_{\phi}$ are just zero, since the original function has only a radial component. Since they go to zero all those partial derivatives go to zero, correct? 
 A: If $U(r)$ is scalar function and you want to show that $\vec F(r) = U(r)\hat r$ has zero curl. Than easiest way to do this is:
$$
\int_0^r U(s) ds = V(r)
$$
$$
U(r) \hat r = V'(r)\hat r  = \nabla V(r)
$$
but
$$
\text{curl } \text{grad } V(r) = 0
$$
A: Fell asleep last night after hitting the "Post" button by accident!  Fixed 'er up first thing this AM, after reading Jesse's corrections.  So here goes the (hopefully) final version:
I'm having a little trouble with your notation since in both the title and at the start of the second paragraph you seem to be referring to a vector field of the form $U(r) \hat r$ (though a parenthesis appears to be missing in the title), where $\hat r = (x, y, z)^T$ is the usual radial vector and $U(r)$ is a spherically symmetric function, i.e., one that only depends on the radius $r = \Vert \hat r \Vert$.  But then when you write out the component formula it's not very clear to me what the $u_r, u_\theta, u_\phi$ refer to, unless perchance it is to the components of $U(r) \hat {\mathbf r}$ in spherical coordinates.  
NEVERTHELESS, if what you really want to do is show that $\nabla \times U(r) \hat r = 0$ for spherically stmmetric functions $U(r)$, then perhaps the easiest thing to do is use the identity
$\nabla \times (f \hat V) = \nabla f \times \hat V + f \nabla \times \hat V, \tag{1}$
a standard result from vector calculus, which can easily be derived from the well-known "determinant formula" for $\nabla \times$:
$\nabla \times \hat Y = \begin{bmatrix} \hat {\mathbf i} & \hat {\mathbf j} & \hat {\mathbf k} \\ \frac{\partial}{\partial x} &  \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat Y_x & \hat Y_y & \hat Y_z \end{bmatrix}. \tag{2}$
If we set $\hat Y = f \hat V$ in (2) and grind, we get
$\nabla \times f \hat V = (\frac{\partial}{\partial y} (f \hat V_z) - (\frac{\partial}{\partial z} (f \hat V_y))\hat {\mathbf i} + (\frac{\partial}{\partial z} (f \hat V_x) - (\frac{\partial}{\partial x} (f \hat V_z))\hat {\mathbf j}$
$+ (\frac{\partial}{\partial x} (f \hat V_y) - (\frac{\partial}{\partial y} (f \hat V_x))\hat {\mathbf k}, \tag{3}$
or
$\nabla \times f \hat V = ((f_y \hat V_z + f \hat V_{z, y}) -  (f_z \hat V_y + f \hat V_{y, z}))\hat {\mathbf i} + ((f_z \hat V_x + f \hat V_{x, z}) - (f_x \hat V_z + f \hat V_{z, x}))\hat {\mathbf j}$
$+ ((f_x \hat V_y + f \hat V_{y, x}) - (f_y \hat V_x + f \hat V_{x, y}))\hat {\mathbf k}, \tag{4}$
after which some simple algebraic re-arrangement yields (1).  For example, the coefficient of $\hat{\mathbf i}$ is
$(f_y \hat V_z - f_z \hat V_y) + f(\hat V_{z, y} - \hat V_{y, z}). \tag{5}$
In (3), (4), and (5), I have used subscript notation for partials, e.g. $f_x = \frac{\partial f}{\partial x}$ and $\hat V_{x,y} = \frac{\partial V_x}{\partial y}$ etc.  This formula may also be found in this wikipedia article.
Applying (1) to $U(r) \hat r$, we have
$\nabla \times U(r) \hat r = \nabla U(r) \times \hat r + U(r) \nabla \times \hat r. \tag{6}$
Now
$\nabla \times \hat r = 0, \tag{7}$
which is easy to see by direct calculation, using (2) if you like, so we are left with
$\nabla \times U(r) \hat r = \nabla U(r) \times \hat r; \tag{8}$
but I claim that for any function $U(r)$ which only depends on $r$, $\nabla U(r)$ is in fact collinear with $\hat r$; to see this, write
$U(r) = U((x^2 + y^2 + z^2)^{\frac{1}{2}}), \tag{9}$
so that for example
$\frac{\partial U}{\partial x}(r) = \frac{dU}{dr}(r) \frac{x}{r}, \tag{10}$
by the chain rule, since $\frac{\partial r}{\partial x} = \frac{x}{r}$.  Similar expressions hold for the $y$ and $z$ derivatives as well.  Now (10), when combined with its $y$ and $z$ counterparts, gives
$\nabla U(r) = r^{-1}\frac{dU}{dr}(r) \hat r, \tag{11}$,
and substituting this in (8) yields
$\nabla \times U(r) \hat r = r^{-1}\frac{dU}{dr}(r) \hat r \times \hat r = 0, \tag{12}$
since $\hat r \times \hat r$ vanishes identically.  Thus we see that
$\nabla \times U(r) \hat r = 0, \tag{13}$
as was required.  QED.
CAVEAT:  One should of course take care applying these or any derivative formula to a function or vector field at the origin $(0, 0, 0)$ where $r = 0$, since $\sqrt{x^2 + y^2 + z^2}$ is continuous but not differentiable there.  I believe similar remarks apply to the answer given by tom.  But if $U(r)$ is sufficiently smooth everywhere, then I do b'lieve she'll fly, Wilbur!
Hope this helps.  Cheers!
and as always
Fiat Lux!
Proving that $\nabla \times (U(r) \hat{r} = 0 $
