[proofplan]
We first verify that $J/I$ is an ideal in $\mathfrak g/I$ by computing brackets of representatives and using that $J$ is an ideal of $\mathfrak g$. We then construct the natural quotient map from the double quotient $(\mathfrak g/I)/(J/I)$ to $\mathfrak g/J$ and check that it is independent of the representative chosen. Finally, we prove that this map is linear, bijective, and bracket-preserving, which gives the desired Lie algebra isomorphism.
[/proofplan]
[step:Show that $J/I$ is an ideal of $\mathfrak g/I$]
Let
\begin{align*}
J/I := \{j + I \in \mathfrak g/I : j \in J\}.
\end{align*}
Since $J$ is a vector subspace of $\mathfrak g$ and $I \subset J$, the set $J/I$ is a vector subspace of $\mathfrak g/I$.
We verify the ideal property. Let $x + I \in \mathfrak g/I$ with $x \in \mathfrak g$, and let $j + I \in J/I$ with $j \in J$. The Lie bracket in the quotient Lie algebra $\mathfrak g/I$ is defined by
\begin{align*}
[x + I, j + I]_{\mathfrak g/I} = [x,j]_{\mathfrak g} + I.
\end{align*}
Because $J$ is an ideal of $\mathfrak g$ and $j \in J$, we have $[x,j]_{\mathfrak g} \in J$. Hence
\begin{align*}
[x + I, j + I]_{\mathfrak g/I} \in J/I.
\end{align*}
Therefore $J/I \trianglelefteq \mathfrak g/I$.
[/step]
[step:Define the natural map from the double quotient]
Define
\begin{align*}
\Phi: (\mathfrak g/I)/(J/I) &\to \mathfrak g/J \\
(x + I) + (J/I) &\mapsto x + J.
\end{align*}
We prove that $\Phi$ is well defined. Suppose
\begin{align*}
(x + I) + (J/I) = (y + I) + (J/I)
\end{align*}
in $(\mathfrak g/I)/(J/I)$, where $x,y \in \mathfrak g$. By equality of cosets in the quotient by $J/I$, this means
\begin{align*}
(x + I) - (y + I) \in J/I.
\end{align*}
Equivalently,
\begin{align*}
(x-y)+I \in J/I.
\end{align*}
Thus there exists $j \in J$ such that
\begin{align*}
(x-y)+I = j+I.
\end{align*}
Therefore $x-y-j \in I$. Since $I \subset J$ and $j \in J$, we get $x-y \in J$. Hence
\begin{align*}
x+J = y+J.
\end{align*}
So $\Phi$ is independent of the chosen representative and is well defined.
[/step]
[step:Prove that the natural map is a linear bijection]
The map $\Phi$ is linear because addition and scalar multiplication in both quotient spaces are defined by representatives. Explicitly, for $x,y \in \mathfrak g$ and $a,b \in k$,
\begin{align*}
\Phi\left(a((x+I)+(J/I)) + b((y+I)+(J/I))\right)
&= \Phi\left((ax+by+I)+(J/I)\right) \\
&= ax+by+J \\
&= a(x+J) + b(y+J) \\
&= a\Phi((x+I)+(J/I)) + b\Phi((y+I)+(J/I)).
\end{align*}
The map $\Phi$ is surjective. If $z+J \in \mathfrak g/J$ with $z \in \mathfrak g$, then
\begin{align*}
\Phi((z+I)+(J/I)) = z+J.
\end{align*}
The map $\Phi$ is injective. Suppose $x \in \mathfrak g$ satisfies
\begin{align*}
\Phi((x+I)+(J/I)) = 0+J.
\end{align*}
Then $x+J = 0+J$, so $x \in J$. Hence $x+I \in J/I$, and therefore
\begin{align*}
(x+I)+(J/I) = 0+(J/I)
\end{align*}
in $(\mathfrak g/I)/(J/I)$. Thus $\ker \Phi = \{0+(J/I)\}$, so $\Phi$ is injective.
[/step]
[step:Check that the natural map preserves Lie brackets]
Let
\begin{align*}
X := (x+I)+(J/I), \qquad Y := (y+I)+(J/I)
\end{align*}
be elements of $(\mathfrak g/I)/(J/I)$, where $x,y \in \mathfrak g$. The Lie bracket in the double quotient is induced from the bracket in $\mathfrak g/I$, so
\begin{align*}
[X,Y]_{(\mathfrak g/I)/(J/I)}
= ([x,y]_{\mathfrak g}+I)+(J/I).
\end{align*}
Applying $\Phi$ gives
\begin{align*}
\Phi([X,Y]_{(\mathfrak g/I)/(J/I)})
= [x,y]_{\mathfrak g}+J.
\end{align*}
On the other hand,
\begin{align*}
[\Phi(X),\Phi(Y)]_{\mathfrak g/J}
= [x+J,y+J]_{\mathfrak g/J}
= [x,y]_{\mathfrak g}+J.
\end{align*}
Therefore
\begin{align*}
\Phi([X,Y]_{(\mathfrak g/I)/(J/I)})
= [\Phi(X),\Phi(Y)]_{\mathfrak g/J}.
\end{align*}
Thus $\Phi$ preserves Lie brackets. Since $\Phi$ is also a linear bijection, it is a Lie algebra isomorphism. Consequently,
\begin{align*}
(\mathfrak g/I)/(J/I) \cong \mathfrak g/J.
\end{align*}
[/step]
