When is a subgroup of a Lie group itself a Lie group? Assume that $G$ is a Lie group. Is it understood which subgroups of $G$ are Lie groups? 
Ideally, I would like to make no extra assumption about $G$. In particular, $G$ can be infinite dimensional.
I am aware that in finite dimensions, Cartan's theorem ensures that any closed subgroup is a Lie group.
In Neeb's notes about infinite dimensional Lie groups, it is mentioned that already for Banach-Lie groups, Cartan's theorem is not true anymore. However locally compact subgroups are Lie subgroups, see p.59.
Is there anything known beyond this?
 A: Consider the situation in finite dimension, and assume that Lie groups are second countable. 
A second countable, locally Euclidean group can have at most one differentiable structure making it into a Lie group. This follows from the fact that a continuous homomorphism between Lie groups is automatically smooth. Now, the condition that a Lie subgroup $H$ of a Lie group $G$ has the induced topology is very restrictive. It implies that $H$ is closed in $G$, so you are in the situation described by Cartan's theorem (essentially one shows that the inclusion map is proper, so it has a closed image), see the book by F. Warner, 3.29.
A Lie group $G$ has no small subgroups (namely, there is a neighborhood of the identity containing no non-trivial subgroups). If $H$ is a subgroup, it follows that it does not contain small subgroups either. Now if $H$ is locally compact with the induced topology, then it follows fom Gleason-Montgomery-Samelson solution to Hilbert's 5th problem that $H$ is a Lie group (and in particular that $H$ is closed in $G$). 
If you relax the condition on the topology of $H$, then $H$ is a Lie subgroup
of $G$ if and only if it is a (second countable) submanifold of $G$. The idea
of the proof is also related to Frobenius; namely, if $H$ is a submanifold and an abstract group, show that it is the leaf of the involutive left-invariant distribution determined by its tangent space at the identity, see Warner, 3.20. 
A: If you define a manifold to be a paracompact Hausdorff space locally homeomorphic to Euclidean space, and a Lie group to be a manifold and a group with continuous group operation, then every subgroup of a Lie group is a Lie subgroup; see Sharpe, Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, p xii. Furthermore, if $G$ is a Lie group equipped with a smooth structure for which the multiplication is a smooth map, then every subgroup of $G$ is a smooth submanifold. 
The point is that we don't ask here for second countability.
