I found myself working on this same problem (for homework), and I think I've written a fairly detailed solution. So I will post it here, in case it is helpful to anyone else.
Let $\mathfrak{g}$ be a 1-dimensional Lie algebra, and let $\{E_1\}$ be a basis for $\mathfrak{g}$. Then for any two vector fields $X,Y\in\mathfrak{g}$, we have $X=aE_1$ and $Y=bE_1$, for some $a,b\in\mathbb{R}$. Thus,
$$[X,Y]=[aE_1,bE_1]=ab[E_1,E_1]=0$$
for all $X,Y\in\mathfrak{g}$. Therefore, the only 1-dimensional Lie algebra is the trivial one. The map
$$\varphi:\mathfrak{g}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$
$$\varphi:aE_1\mapsto
\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)$$
is a Lie algebra homomorphism, since
$$\varphi([aE_1,bE_1])=\varphi(0)=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right)\mbox{, and}$$
$$[\varphi(aE_1),\varphi(bE_1)]=\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)\left(\begin{array}{ll}
b&0\\
0&0
\end{array}\right)-\left(\begin{array}{ll}
b&0\\
0&0
\end{array}\right)\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)$$
$$=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right).$$
Thus, $\mathfrak{g}$ is isomorphic to the (abelian) Lie subalgebra
$$\varphi(\mathfrak{g})=\left\{\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
Now let $\mathfrak{h}$ be a 2-dimensional Lie algebra, and let $\{E_1,E_2\}$ be a basis for $\mathfrak{h}$. Then for any two vector fields $X,Y\in\mathfrak{h}$, we have $X=aE_1+bE_2$ and $Y=cE_1+dE_2$, for some $a,b,c,d\in\mathbb{R}$. Thus,
$$\begin{array}{ll}
[X,Y]&=[aE_1+bE_2,cE_1+dE_2]\\
&=a[E_1,cE_1+dE_2]+b[E_2,cE_1+dE_2]\\
&=ac[E_1,E_1]+ad[E_1,E_2]+bc[E_2,E_1]+bd[E_2,E_2]\\
&=(ad-bc)[E_1,E_2].
\end{array}$$
If $[E_1,E_2]=0$, then we have the trivial 2-dimensional Lie algebra. The map
$$\varphi:\mathfrak{h}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$
$$\varphi:aE_1+bE_2\mapsto
\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)$$
is a Lie algebra homomorphism, since
$$\varphi([aE_1+bE_2,cE_1+dE_2])=\varphi(0)=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right)\mbox{, and}$$
$$[\varphi(aE_1+bE_2),\varphi(cE_1+dE_2)]=\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)\left(\begin{array}{ll}
c&0\\
0&d
\end{array}\right)-\left(\begin{array}{ll}
c&0\\
0&d
\end{array}\right)\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)$$
$$=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right).$$Furthermore, this map is faithful (injective). Thus, $\mathfrak{h}$ is isomorphic to the (abelian) Lie subalgebra
$$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
If $[E_1,E_2]\neq0$, then set $E_3=[E_1,E_2]$. Then for all $X,Y\in\mathfrak{h}$ we have $[X,Y]=\lambda E_3$ for some $\lambda\in\mathbb{R}$. In particular, for any $E_4\in\mathfrak{g}$ such that $E_4$ and $E_3$ are linearly independent, we have $[E_4,E_3]=\lambda_0 E_3$. Replacing $E_4$ with $1/\lambda_0 E_4$, we now have a basis $\{E_4, E_3\}$ for $\mathfrak{g}$ such that $[E_4, E_3]=E_3$. The map
$$\varphi:\mathfrak{h}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$
$$\varphi:aE_4+bE_3\mapsto
\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)$$
is a Lie algebra homomorphism, since
$$\varphi([aE_4+bE_3,cE_4+dE_3])=\varphi((ad-bc)E_3)=\left(\begin{array}{ll}
0&ad-bc\\
0&0
\end{array}\right)\mbox{, and}$$
$$[\varphi(aE_4+bE_3),\varphi(cE_4+dE_3)]=\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)\left(\begin{array}{ll}
c&d\\
0&0
\end{array}\right)-\left(\begin{array}{ll}
c&d\\
0&0
\end{array}\right)\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)$$
$$=\left(\begin{array}{ll}
0&ad-bc\\
0&0
\end{array}\right).$$Furthermore, this map is faithful (injective). Thus, $\mathfrak{h}$ is isomorphic to the (non-abelian) Lie subalgebra
$$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
Best Answer
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 $\mathfrak{so}(3)$.