Consider four dimensional Lie algebra with non-zero commutations:
$[e_{2},e_{3}]=e_{1}, [e_{2}, e_{4}]=e_{2}, [e_{3}, e_{4}]=-e_{3}$
having sub-algebras $S_{1}=\{e_{1}, e_{2}\}, S_{2}=\{e_{1}+e_{2}\}, S_{3}=\{e_{1}+e_{4}, e_{3}\}, S_{4}=\{e_{3}\}$.
In terminology of sub-sets as defined in general set theory, how $S_{1}$ and $S_{2}$ are comparable ?
Is $S_{4}$ is sub-set of $S_{3}$ ?
Note that subsets are not enough here, since we are dealing with vector spaces. So we need subspaces. A Lie subalgebra is a linear subspace which is closed under the Lie bracket. In your example, $S_1=span(e_1,e_2)$ is an abelian subalgebra of dimension $2$.