[proofplan] We prove directly from the definition of an ideal. First we note that $I^\perp$ is a linear subspace of $\mathfrak g$ because it is defined by homogeneous linear equations. Then, for $x \in I^\perp$ and $g \in \mathfrak g$, we test the bracket $[g,x]$ against an arbitrary element $i \in I$. Invariance and symmetry move the bracket from $[g,x]$ onto $[i,g]$, which lies in $I$ because $I$ is an ideal; orthogonality of $x$ to $I$ then gives the desired vanishing. [/proofplan]

[step:Verify that $I^\perp$ is a linear subspace of $\mathfrak g$] Let $x,y \in I^\perp$ and let $a,b \in k$. For every $i \in I$, bilinearity of $B$ gives \begin{align*} B(ax+by,i)=aB(x,i)+bB(y,i)=a \cdot 0+b \cdot 0=0. \end{align*} Therefore $ax+by \in I^\perp$, so $I^\perp$ is a linear subspace of $\mathfrak g$. [/step]

[step:Show that bracketing an element of $I^\perp$ by an element of $\mathfrak g$ stays orthogonal to $I$] Let $x \in I^\perp$ and let $g \in \mathfrak g$. Define \begin{align*} y := [g,x] \in \mathfrak g. \end{align*} To prove $y \in I^\perp$, let $i \in I$ be arbitrary. By symmetry of $B$, followed by invariance of $B$ applied to the triple $(x,i,g)$, we have \begin{align*} B(y,i) &= B([g,x],i) \\ &= B(i,[g,x]) \\ &= B([i,g],x) \\ &= B(x,[i,g]). \end{align*} Since $I \trianglelefteq \mathfrak g$ and $i \in I$, $g \in \mathfrak g$, the ideal property gives $[i,g] \in I$. Because $x \in I^\perp$, it follows that \begin{align*} B(x,[i,g])=0. \end{align*} Hence $B(y,i)=0$ for every $i \in I$, so $y=[g,x] \in I^\perp$. [/step]

[step:Conclude that $I^\perp$ is an ideal] We have shown that $I^\perp$ is a linear subspace of $\mathfrak g$ and that \begin{align*} [g,x] \in I^\perp \end{align*} for every $g \in \mathfrak g$ and every $x \in I^\perp$. This is precisely the condition $[\mathfrak g,I^\perp] \subseteq I^\perp$, so $I^\perp \trianglelefteq \mathfrak g$. [/step]