[proofplan]
We compare the derived series of $\mathfrak a$, $\mathfrak g$, and $\mathfrak g/\mathfrak a$. In one direction, the derived series of a Lie subalgebra is contained termwise in the derived series of the ambient Lie algebra, and the derived series of a quotient is the image of the derived series of the original algebra. In the converse direction, solvability of $\mathfrak g/\mathfrak a$ forces a sufficiently deep derived algebra of $\mathfrak g$ to lie inside $\mathfrak a$, and solvability of $\mathfrak a$ then kills the remaining derived terms.
[/proofplan]
[step:Define the derived series and record its monotonicity]
For any Lie algebra $\mathfrak h$ over $F$, define its derived series $(\mathfrak h^{(m)})_{m \geq 0}$ by
\begin{align*}
\mathfrak h^{(0)} &= \mathfrak h, &
\mathfrak h^{(m+1)} &= [\mathfrak h^{(m)}, \mathfrak h^{(m)}]
\end{align*}
for every integer $m \geq 0$. The Lie algebra $\mathfrak h$ is solvable precisely when there exists an integer $m \geq 0$ such that $\mathfrak h^{(m)} = 0$.
If $\mathfrak h \subset \mathfrak k$ is a Lie subalgebra, then
\begin{align*}
\mathfrak h^{(m)} \subset \mathfrak k^{(m)}
\end{align*}
for every $m \geq 0$. This follows by induction on $m$: the case $m = 0$ is the inclusion $\mathfrak h \subset \mathfrak k$, and if $\mathfrak h^{(m)} \subset \mathfrak k^{(m)}$, then
\begin{align*}
\mathfrak h^{(m+1)}
= [\mathfrak h^{(m)}, \mathfrak h^{(m)}]
\subset [\mathfrak k^{(m)}, \mathfrak k^{(m)}]
= \mathfrak k^{(m+1)}.
\end{align*}
[/step]
[step:Show that solvability descends to the ideal and the quotient]
Assume that $\mathfrak g$ is solvable. Choose an integer $N \geq 0$ such that $\mathfrak g^{(N)} = 0$.
Since $\mathfrak a \subset \mathfrak g$ is a Lie subalgebra, the monotonicity proved above gives
\begin{align*}
\mathfrak a^{(N)} \subset \mathfrak g^{(N)} = 0.
\end{align*}
Thus $\mathfrak a$ is solvable.
Now let
\begin{align*}
\pi: \mathfrak g &\to \mathfrak g/\mathfrak a \\
x &\mapsto x + \mathfrak a
\end{align*}
be the quotient Lie algebra homomorphism. We claim that
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m)} = \pi(\mathfrak g^{(m)})
\end{align*}
for every $m \geq 0$. For $m = 0$, this is the surjectivity of $\pi$. If the identity holds for $m$, then
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m+1)}
&= [(\mathfrak g/\mathfrak a)^{(m)},(\mathfrak g/\mathfrak a)^{(m)}] \\
&= [\pi(\mathfrak g^{(m)}),\pi(\mathfrak g^{(m)})] \\
&= \pi([\mathfrak g^{(m)},\mathfrak g^{(m)}]) \\
&= \pi(\mathfrak g^{(m+1)}),
\end{align*}
because $\pi$ preserves Lie brackets. Therefore
\begin{align*}
(\mathfrak g/\mathfrak a)^{(N)} = \pi(\mathfrak g^{(N)}) = \pi(0) = 0,
\end{align*}
so $\mathfrak g/\mathfrak a$ is solvable.
[/step]
[step:Lift solvability from the quotient and finish inside the ideal]
Conversely, assume that $\mathfrak a$ and $\mathfrak g/\mathfrak a$ are solvable. Choose integers $R,S \geq 0$ such that
\begin{align*}
(\mathfrak g/\mathfrak a)^{(R)} &= 0, &
\mathfrak a^{(S)} &= 0.
\end{align*}
Using the quotient identity from the previous step, we have
\begin{align*}
0 = (\mathfrak g/\mathfrak a)^{(R)} = \pi(\mathfrak g^{(R)}).
\end{align*}
Thus every element of $\mathfrak g^{(R)}$ lies in $\ker \pi = \mathfrak a$, so
\begin{align*}
\mathfrak g^{(R)} \subset \mathfrak a.
\end{align*}
Applying monotonicity to the inclusion $\mathfrak g^{(R)} \subset \mathfrak a$ gives
\begin{align*}
(\mathfrak g^{(R)})^{(S)} \subset \mathfrak a^{(S)} = 0.
\end{align*}
By the recursive definition of the derived series,
\begin{align*}
(\mathfrak g^{(R)})^{(S)} = \mathfrak g^{(R+S)}.
\end{align*}
Hence
\begin{align*}
\mathfrak g^{(R+S)} = 0.
\end{align*}
Therefore $\mathfrak g$ is solvable.
[/step]
[step:Combine the two implications]
We have proved that if $\mathfrak g$ is solvable, then both $\mathfrak a$ and $\mathfrak g/\mathfrak a$ are solvable. We have also proved that if both $\mathfrak a$ and $\mathfrak g/\mathfrak a$ are solvable, then $\mathfrak g$ is solvable. These two implications establish the stated equivalence.
[/step]
