The new question amounts to classifying, up to isomorphism, 3-dimensional subalgebras of the real Lie algebra $\mathfrak{su}(2,1)$.

If I'm correct, the answer is:

- simple split subalgebras (isomorphic to $\mathfrak{sl}_2$).
- simple non-split subalgebras (isomorphic to $\mathfrak{so}_3$).
- Heisenberg subalgebras
- subalgebras with basis $(T,X,Y)$ with $[T,X]=X$, $[T,Y]=2Y$, $[X,Y]=0$.

In particular, this discards the Lie algebra with basis $(T,X,Y)$, $[T,X]=X$, $[T,Y]=\lambda Y$, $[X,Y]=0$ whenever $\lambda\notin\{2,\frac12\}$ (the one written as an example by the OP is $\lambda=1$).

To prove this: the first part is to realize each of these algebras. The first two appear as $\mathfrak{su}(1,1)$ and $\mathfrak{su}(2)$. The last two will pop up in the proof.

Consider the action of a 3-dimensional subalgebra $\mathfrak{h}$ on $\mathbf{C}^3$. If it is irreducible, $\mathfrak{h}$ is semisimple and this falls in one of the first two cases.

So it preserves a line or plane, and preserves its orthogonal too, so preserves a line. If the line is non-isotropic, its orthogonal is a stable supplement, and thus $\mathfrak{h}$ falls into $\mathfrak{s}(\mathfrak{u}(1)\times\mathfrak{u}(2))$ or $\mathfrak{s}(\mathfrak{u}(1)\times\mathfrak{u}(1,1))$, which are 4-dimensional and isomorphic to a product of an abelian 1-dimensional Lie algebra and a simple 3-dimensional Lie algebra. This forces $\mathfrak{h}$ to be simple, and this case is done (of course, if we classify up to conjugation, this is a distinct case! but we don't.)

The last case is when it preserves an isotropic line. In this case let us say that the hermitian form is given by the matrix $\begin{pmatrix} 0 &0&1\\0&1&0\\1&0&0\end{pmatrix}$. Then $\mathfrak{u}(2,1)$ consists of the matrices $$\begin{pmatrix} x&y&it\\z&iu&-\bar{y}\\iv&-\bar{z}&-\bar{x}\end{pmatrix}\quad x,y,z\in\mathbf{C},\;t,u,v\in\mathbf{R},$$
that is, which are skew-hermitian with respect to the antidiagonal; $\mathfrak{u}(2,1)$ consists of those trace zero matrices therein, i.e., for which $iu=-x+\bar{x}$. Since all isotropic vectors are in the same orbit, it is enough to consider the case when $\mathfrak{h}$ preserves the line $(\mathbf{C},0,0)$, that is, is contained in the 5-dimensional subalgebra $\mathfrak{w}$ consisting of those matrices
$$\begin{pmatrix} x&y&it\\0&-x+\bar{x}&-\bar{y}\\0&0&-\bar{x}\end{pmatrix}\quad x,y\in\mathbf{C},\;t\in\mathbf{R}.$$

So we have to classify those 3-dimensional subalgebras of $\mathfrak{w}$ up to isomorphism. Let $\mathfrak{h}$ be solvable and 3-dimensional. We discuss on the dimension $D\in\{0,1,2\}$ of $[\mathfrak{h},\mathfrak{h}]$.

The following facts hold for an arbitrary solvable 3-dimensional Lie algebra over a field of characteristic zero (and certainly also all characteristic except maybe 2,3):
$D=0$ iff $\mathfrak{h}$ is abelian, $D=1$ iff $\mathfrak{h}$ is Heisenberg or product of a 1-dimensional abelian with the non-abelian 2-dimensional Lie algebra (according to whether the 1-dimensional derived subalgebra is central or not). So we have to discard the abelian case, and to treat the cases when $D=2$.

For the abelian case, observe that for a matrix as above, if $x\neq 0$ it is diagonalizable with distinct eigenvalues; then its centralizer is diagonalizable but is constrained to have trace zero, so has dimension $\le 2$. The for $x=0$ what remains precisely consists of a Heisenberg subalgebra.

Let us discard the non-abelian product case; it has 1-dimensional center. Consider a matrix as above: as in the abelian case, if $x\neq 0$, it has abelian centralizer, which is now excluded. So $x=0$. If $x=0$ and $y\neq 0$, this is conjugate to a Jordan nilpotent matrix and again has abelian centralizer. So the center is the line generated by $E_{13}$. So $\mathfrak{h}$ is contained in the centralizer in $\mathfrak{w}$ of $E_{13}$ which consists of those matrices as above with $x$ purely imaginary, namely those
$$\begin{pmatrix} is&y&it\\0&-2is&-\bar{y}\\0&0&is\end{pmatrix}\quad y\in\mathbf{C},\;t,s\in\mathbf{R}.$$
This 4-dimensional Lie algebra has a basis $T,X,Y,Z$ with $Z$ central and $[T,X]=Y$, $[T,Y]=-X$, $[X,Y]=Z$. It is easy to check that its only 3-dimensional subalgebra is the Heisenberg one with basis $(X,Y,Z)$ (here we use that we work with reals). So this case is excluded.

Finally let us treat the $D=2$ case. Then $[\mathfrak{h},\mathfrak{h}]$ is contained in $[\mathfrak{w},\mathfrak{w}]$ (which is Heisenberg), and is abelian. In a Heisenberg Lie algebra, the 2-dimensional subalgebras consists of those planes containing the center. Hence $[\mathfrak{h},\mathfrak{h}]$ consists, for some nonzero complex number $y_0$, of those

$$\begin{pmatrix} 0&\lambda y_0&it\\0&0&-\lambda\bar{y_0}\\0&0&0\end{pmatrix}\quad t,\lambda\in\mathbf{R}.$$
Then $\mathfrak{h}$ is contained in the normalizer of this 2-dimensional subalgebra in $\mathfrak{w}$, which consists of those matrices as above with $x$ real, namely
$$\begin{pmatrix} x&y&it\\0&0&-\bar{y}\\0&0&-x\end{pmatrix}\quad y\in\mathbf{C},\;x,t\in\mathbf{R}.$$
So $\mathfrak{h}$ has a basis $(T,X,Y)$, with
$$T=\begin{pmatrix} 1&y&it\\0&0&-\bar{y}\\0&0&-1\end{pmatrix},X=\begin{pmatrix} 0&y_0&0\\0&0&-\bar{y_0}\\0&0&0\end{pmatrix},Y=\begin{pmatrix} 0&0&i\\0&0&0\\0&0&0\end{pmatrix},$$
which indeed satisfy $[T,X]=X,[T,Y]=2Y$, $[X,Y]=0$.

injectivehomomorphism. Injective means that some minor is nonzero. This is why I mentioned "constructible subset", not just a Zariski closed subset. Btw all these methods are valid with arbitrary algebras, Lie axioms play no role. $\endgroup$ – YCor Nov 28 '17 at 6:25