Are there indecomposable unsolvable four and five dimensional Lie algebras?

Every finite dimensional real Lie algebra has the form $\mathfrak g=\mathfrak l\ltimes\mathfrak r$ where $\mathfrak l$ is semisimple and $\mathfrak r$ is solvable (Levi decomposition). The smallest semisimple Lie algebras are $\mathfrak{su}(1,1)=\mathfrak{sl}(2,\mathbb R)$ and $\mathfrak{su}(2)$, both of dimension $3$. The next larger ones have dimension $6$. So if $\mathfrak g$ is non-solvable of dimension $\le5$ then $\mathfrak l$ is one of the two $3$-dimensional ones. The indecomposability implies that $\mathfrak l$ acts non-trivially on $\mathfrak r$. From this, one gets easily that $\mathfrak{su}(1,1)$, $\mathfrak{su}(2)$, and $\mathfrak{sl}(2,\mathbb R)\ltimes\mathbb R^2$ are all the non-solvable indecomposable Lie algebras of dimension $\le5$. The last algebra is item $A_{5,40}$ of Table II of loc. cit. which is incorrectly labeled as solvable.

As already mentioned, an easy argument using the Levi decomposition shows that there are three distinct indecomposable non-solvable Lie algebras of dimension $5$.
However, I think it is worth to point out in this context that a full classification of all real $5$-dimensional Lie algebras has been given by G.M. Mubarakzyanov in $1963$, Izv. Vyssh. Uchebn. Zaved. Mat. 34. Besides the indecomposable non-solvable algebras we also have of course the decomposable non-solvable Lie algebras $\mathfrak{sl}(2,\mathbb R)\times \mathfrak{r}_2(\mathbb{R})$ and $\mathfrak{sl}(2,\mathbb R)\times \mathbb{R}^2$, where $\mathfrak{r}_2(\mathbb{R})$ denotes the non-abelian Lie algebra of dimension $2$.
The Lie brackets for $\mathfrak{sl}(2,\mathbb R)\ltimes\mathbb R^2$, in a basis $e_1,\ldots e_5$ are given by $$ [e_1,e_2]=e_3,\; [e_1,e_3]=-2e_1,\; [e_1,e_5]=e_4, \; [e_2,e_3]=2e_2,\; [e_2,e_4]=e_5, $$

$$ [e_3,e_4]=e_4,\; [e_3,e_5]=-e_5. $$

A further reference is the paper of Roman Popovych (2003).

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 $\mathfrak{su}_2\ltimes\mathbf{C}^2$.