I am not familiar with the theory of Lie groups, so I am having a hard time finding all of the connected closed real Lie subgroups of $\mathrm{SL}(2, \mathbb{C})$, up to conjugation.
One can find the real and complex parabolic, elliptic, hyperbolic, subgroups, $\mathrm{SU}(2)$, $\mathrm{SU}(1,1)$ and $\mathrm{SL}(2,\mathbb{R})$ (the last two ones are isomorphic however), the subgroup of real upper triangular matrices, the subgroup of upper triangular matrices with unitary diagonal coefficients, and the subgroup of complex triangular matrices.
Are there any others?
I will start with the general strategy and the will show how to use it in the case of $SL(2,C)$. Let $G$ be a connected Lie subgroup of another Lie group $H$; in your case, $H$ is a complex Lie group, which helps. You first look at the Levi-Malcev decomposition of the Lie algebra $G$ and ask if the solvable radical is nontrivial. If it is, then its exponential is a normal solvable Lie subgroup $S$ of $H$ and, hence, it has a fixed for any action on a finite-dimensional complex projective space. The set of such fixed vectors will be invariant under $G$.
In your setting, $H$ and, hence, $H$, is acting on $CP^1$, hence, if $S<G$ is nontrivial, then $S$ fixes a point in $CP^1$. Dimension of the fixed-point set has to be zero (otherwise $S$ acts trivially on $CP^1$ which is impossible) and you trivially conclude that it is either one or two points. Since $G$ is connected, both points have to be fixed by $G$. Hence, up to conjugation, $G$ is contained in the group $B$ of upper-triangular matrices, a Borel subgroup of $SL(2,C)$ which is solvable (in particular, $G=S$). Now, you have to classify connected subgroups $G$ of $B$. This is not too hard, since you have to ask yourself how do they intersect the commutator subgroup $U$ (consisting of unipotent elements) of $B$. If the intersection is trivial, then $G$ (up to conjugation) embeds in the diagonal subgroup, isomorphic to $C^\times$ and, hence, $G$ is either ${\mathbb C}^\times$ or ${\mathbb R}_+$. If the intersection with $U$ is nontrivial, then you conclude that it is either real 1-dimensional or complex 1-dimensional. Then you see that $G$ is one of the following subgroups sitting naturally in $B$:
${\mathbb R}$ or ${\mathbb C}$ (contained in $U$).
${\mathbb R}\rtimes {\mathbb R}_+$.
${\mathbb C}\rtimes {\mathbb R}_+$.
${\mathbb C}\rtimes {\mathbb S}^1$.
The entire group $B$.
Now, let's consider the more interesting case, when $G$ has semisimple Lie algebra. The thing to do is to look for maximal semisimple subalgebras. In general, there is a classical work by Dynkin from 1950s where he classified maximal semisimple Lie subalgebras (and subgroups) of semisimple Lie groups.
In your case, everything again can be done by hand. First of all, $G$ has to have (real) rank $\le 1$ (since $SL(2, C)$ has rank 1) and, hence, be simple (in the sense that its Lie algebra is simple); moreover, real dimension of $G$ is at most 6. If you think in terms of classification of simple Lie algebras, this leaves you with only two options: $su(2)$, $sl(2,R)$ and $sl(2,C)$. Now, in the $su(2)$ case, the group $G$ has to be compact (since its Lie algebra is "compact"); hence, by Cartan's theorem, is is contained (up to conjugation) in the maximal compact subgroup $SU(2)< SL(2,C)$. If the Lie algebra is $sl(2,C)$ then the corresponding Lie groups is the entire $SL(2,C)$. Lastly, in the case of $sl(2,R)$, with a bit more work you see that $G$ is conjugate to the set of real points $SL(2,R)< SL(2,C)$. (First verify this on the level of Lie algebras.)