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 nbhd 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 relaret 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.

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.