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 upshot is that Table II of loc.cit. has in fact a gap.

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.