The classification of indecomposable non-solvable Lie algebras in dimension $\le 7$ over a field of characteristic zero is a simple exercise.
First, the semisimple ones have dimension 3 or 6: in dimension 3, $\mathfrak{so}(q)$ for quadratic forms in 3 variables, in dimension 6, the same over a quadratic extension, or products of two simple ones of dimension 3.
Next we use a Levi decomposition $\mathfrak{s}\ltimes\mathfrak{r}$. If $\mathfrak{s}$ has dimension 6, $\mathfrak{r}$ has dimension $\le 1$ and is a direct factor.
If $\mathfrak{s}$ has dimension 3, one has to discuss on $\mathfrak{r}$. If $\mathfrak{r}$ is abelian, it comes with a representation of $\mathfrak{r}$, which splits into irreducibles. If a 1-dimensional rep occurs, then it is a direct factor, so in the indecomposable cases the only possibilities are 2,3,4 and 2+2. (Note that if $\mathfrak{s}$ has a nontrivial 2-dimensional rep, then it is isomorphic to $\mathfrak{sl}_2$; the only 3-dimensional irreducible is the adjoint rep. In dim 4, I'm not sure in general fields, but in the real case both the split form and the non-split form have a unique 4-dimensional irreducible rep, which in the case of $\mathfrak{su}_2$ is not absolutely irreducible).
If $\mathfrak{r}$ is nilpotent and not abelian then it is (by the classification of nilpotent Lie algebras of dimension $\le 4$, either $\mathfrak{h}_3$ (Heisenberg), or $\mathfrak{h}_3\times\mathfrak{a}_1$ (its product with a 1-dimensional abelian) or $\mathfrak{f}_4$ (filiform of dimension 4). In the last case, the derivation algebra is solvable and hence $\mathfrak{r}$ is a direct factor. In the first two cases, either $\mathfrak{r}$ is a direct factor, or we have the following: modulo the center, we have an irreducible 2-dimensional $\mathfrak{s}$-module (hence $\mathfrak{s}$ is $\mathfrak{sl}_2$); then this lifts to an irreducible 2-dimensional submodule in $\mathfrak{r}$, and we see we have the (unique) nontrivial semidirect product $\mathfrak{sl}_2\ltimes\mathfrak{h}_3$, or its direct product with $\mathfrak{a}_1$.
The last case is when $\mathfrak{r}$ is not nilpotent. In a solvable Lie algebra $\mathfrak{r}$ with nilpotent radical, it is easy to check that $\dim(\mathfrak{n})\ge\dim(\mathfrak{r})/2$, with equality only for powers of the 2-dimensional nonabelian Lie algebra. The latter has a solvable derivation algebra and hence only occurs as direct factor. Otherwise in our case the pair $(\dim(\mathfrak{r}),\dim(\mathfrak{n}))$ is $(4,3)$ or $(3,2)$. Since in these cases $\mathfrak{n}$ has codimension 1, it has a $\mathfrak{s}$-invariant complement. So the Lie algebra has the form $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes\mathfrak{n}$, where $\mathfrak{a}_1$ does not act on $\mathfrak{n}$ in a nilpotent way, and acts in the centralizer of the $\mathfrak{s}$-action. Excluding the case with abelian direct factors, we find $\mathfrak{gl}_2\ltimes\mathfrak{h}_3$,
$(\mathfrak{s}\times\mathfrak{a}_1)\ltimes\mathfrak{v}$ where
$\mathfrak{v}$ is an irreducible of dimension 2 or 3 and
$\mathfrak{a}_1$ acts by scalar multiplication.
So the list (dimension in parentheses):
- (3) $\mathfrak{s}$ simple of dimension 3 ($\mathfrak{so}$ of some 3-dimensional quadratic form, unique up to scalar multiplication and equivalence: in the real case, $\mathfrak{sl}_2=\mathfrak{so}(2,1)$ or $\mathfrak{su}_2=\mathfrak{so}(3)$)
- (6) $\mathfrak{s}$ simple of dimension 6 (idem over a quadratic extension of the ground field; in the real case: $\mathfrak{sl}_2(\mathbf{C})=\mathfrak{so}_3(\mathbf{C})$)
- (5) $\mathfrak{sl}_2(K)\ltimes K^2$
- (7) $\mathfrak{sl}_2(K)\ltimes (K^2\oplus K^2)$
- (6) $\mathfrak{sl}_2\ltimes\mathfrak{h}_3$
- (6) $\mathfrak{gl}_2(K)\ltimes K^2$
- (7) $\mathfrak{gl}_2\ltimes\mathfrak{h}_3$
- (6) $\mathfrak{s}\ltimes V_\mathfrak{s}$ where $\mathfrak{s}$ is simple of dimension 3 and $V_\mathfrak{s}$ is the adjoint representation of $\mathfrak{s}$
- (7) $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes V_\mathfrak{s}$ where $\mathfrak{s}$ is 3-dimensional simple, $\mathfrak{a}_1$ acts by scalar multiplication
- (7) $\mathfrak{s}\ltimes V$, $V$ irreducible of dimension 4. To be more precise, in the real case, the only case beyond the split case $\mathfrak{sl}_2(K)\ltimes\mathrm{Sym}^3(K^2)$ is $\mathfrak{su}_2\ltimes\mathbf{C}^2$.
