Why is Lie's Third Theorem difficult?

This question is related to the following rather deep result: if $G$ is a finite dimensional Lie group, then $\pi_2(G)=0$.

The analogous statement fails for Lie algebroids, as does Lie III. The paper of Tseng and Zhu shows that this is no accident.

Lie III may be interpreted as saying the following: the foliation of the space of $\frak{g}$-connections on the 1-simplex associated to infinitesimal gauge transformation has a Hausdorff leaf space.

A connection is a one-form $A\in\mathcal{A}=\Omega^1([0,1],\frak{g})$ with values in the Lie algebra $\frak{g}$. The infinitesimal gauge action is given by the formula


where $X\in\Omega^0([0,1],\frak{g})$: these vectors span an integrable distribution in the tangent space of $\mathcal{A}$.

(This leaf space is then the simply connected Lie group associated to $G$.) I was taught this point of view by Raoul Bott.

That gluing together of group chunks, constructed from the BCH formula is precisely more or less what Serre does to prove the theorem (in the first proof he gives in) his book on Lie groups and Lie algebras. [Serre, Jean-Pierre. Lie algebras and Lie groups. 1964 lectures given at Harvard University. Second edition. Lecture Notes in Mathematics, 1500. Springer-Verlag, Berlin, 1992. viii+168 pp.]

There are Lie algebras which are not the Lie algebra of an algebraic group. (See here.) That rules out some proofs, but it doesn't seem like it would be a problem for the sort of analytic proof you are proposing.