Here is a nice example. 

**Theorem.** Space $\newcommand\R{\mathbb{R}}\R^3$ is the disjoint union of circles. 

The AC proof is simply to enumerate all points in $\R^3$ in a well-ordered sequence in length continuum, and then attach a circle to each point along the way. If at stage $\alpha$ the $\alpha$th point is not yet occuring on a circle, then since we have chosen already only fewer than continuum many circles, there must be a plane through the point that is not coplanar with any of them. And since each circle already chosen intersects that plane in at most two points, there are fewer than continuum many forbidden points on the plane. And so we can find a circle through that point in that plane that avoids them.

Meanwhile, there is a constructive proof given as theorem 1.1 of 

 - M. Jonsson and J. Wästlund: Partition of R3 into curves, Mathematica Scandinavica 83 (1998) 192-204; [JSTOR](https://www.jstor.org/stable/24493132), [author's website](http://www.math.chalmers.se/~wastlund/).

One places circles of radius 1 centered at (4k+1,0,0), and then considers spheres centered at the origin, which intersects that family always in exactly 2 points, and such a 2-punctured sphere can be partitioned by circles. 

Of course it is nice to have the specific construction, but to my way of thinking, the AC proof is far more flexible, and one can use to to construct many other kinds of partitions (into unit circles, shapes of other kinds etc etc), whereas specific constructions for many of these other instances remain as open questions.