 Solution

If $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{\beta }$ and $h\in \mathfrak{h}$, invariance of the Killing form givesUnless $\alpha +\beta =0$, some $h$ makes its coefficient nonzero, so the two root spaces are orthogonal. Similarly $B(\mathfrak{h},{\mathfrak{g}}_{\alpha })=0$ for every root: choose $h$ with $\alpha (h)\ne 0$ and use that $\mathfrak{h}$ is abelian.

Now take nonzero $x\in {\mathfrak{g}}_{\alpha }$. Nondegeneracy on $\mathfrak{g}$ supplies some element pairing nontrivially with $x$. Its root-space decomposition shows that only its ${\mathfrak{g}}_{-\alpha }$ component can contribute. Thus the pairing between opposite root spaces is nonzero, and indeed nondegenerate. ThereforeHere nonzero means the bilinear pairing is not identically zero, not that every pair of vectors has a nonzero value.

If $h_0\in \mathfrak{h}$ is orthogonal to all of $\mathfrak{h}$, it is also orthogonal to every root space, by the preceding orthogonality. It is therefore orthogonal to all of $\mathfrak{g}$, forcing $h_0=0$. Hence $B{|}_{\mathfrak{h}}$ is nondegenerate.