So, the question is to find all connected subgroups of $SL(2,\mathbb{R})$. I understand, how to find all closed connected subgroups: they are in one-to-one correspondence with Lie subalgebras of $\mathfrak{sl}(2,\mathbb{R})$. Up to conjugation, they are: orthogonal, diagonal, upper triangular and upper unitriangular matrices. I also know, how to find all path-connected subgroups: the list is the same as in the case of closed subgroups. But I don't know, whether or not it exhausts all connected subgroups.
EDIT: To clarify, the question can be formulated, as whether it is true, that for subgroups of $SL(2,\mathbb{R})$ connectedness and path-connectedness is the same property.
Your question is related to the existence of connected but not path-connected subgroups of $SL_2(\Bbb R)$. I am not sure that there exists a list of all connected subgroups of $SL_2(\Bbb R)$. But it seems that one example showing that connectedness and path-connectedness is not the same property for subgroups of Lie groups is enough (see your edit). There is already a long discussion about the construction of connected, not path-connected subgroups of Lie groups here, with links to MO-posts. The construction is given for the $2$-torus. A similar construction could work for $SL_2(\Bbb R)$.
The answer given here mentiones a Theorem by Thomas:
Theorem (Thomas): Every Lie group of dimension greater than one always admits a connected subgroup which is not a Lie subgroup (and hence isn't path connected, since connected Lie subgroups are path-connected).
However, it is not clear if the proof is entirely correct. Moreover, his definition of a Lie subgroup is different from other definitions (Thomas calls it an "analytic" subgroup).