Lie algebras containing $ so_3(\mathbb{R}) $ Every subalgebra of a solvable Lie algebra is solvable. So a Lie algebra $ \mathfrak{g} $ containing a subalgebra isomorphic to $ \mathfrak{so}_3(\mathbb{R}) $ cannot be solvable.
Is the converse true? In other words, is it true that a Lie algebra $ \mathfrak{g} $ contains a subalgebra isomorphic to $ \mathfrak{so}_3(\mathbb{R}) $ if and only if $ \mathfrak{g} $ is not solvable?

As coudy already said, the answer to your question is no, and as pointed out in the comments by Moishe Kohan, if you modify your question to also consider $\mathfrak{sl}(2,\mathbb{R})$ then the answer becomes yes. So we have
Theorem: A real finite-dimensional Lie algebra is not solvable if and only if it has a subalgebra isomorphism to either $\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{so}(3)$.
Here's a proof sketch: You already proved one direction. For the other direction, note that by the Levi decomposition, every real finite-dimensional Lie algebra is a semidirect product of a solvable Lie algebra and a semisimple Lie algebra. But every real semisimple Lie algebra has a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{so}(3)$.
You can complete the proof sketch without the full power of the classification. For example, the complexification of your real semisimple Lie algebra will be a complex semisimple Lie algebra, and this complex semisimple Lie algebra will have a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{C})$ which is preserved by the complex conjugation map. So your original real semisimple Lie algebra will contain a real form of $\mathfrak{sl}(2,\mathbb{C})$, and there are only two of these: $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{so}(3)$.
