For real Lie algebras, is any invariant bilinear form a scalar multiple of the Killing form?

I know that for a complex Lie algebra, Schur's lemma can be used to show that any invariant bilinear form on a simple Lie algebra is a scalar multiple of the killing form, but Schur's lemma does not hold for $\mathbb{R}$, so I guess this isn't true for real simple Lie algebras. Does someone know a example? I can't seem to find one.

Actually for a Lie algebra $\mathfrak{g}$ over a field $K$ and a field extension $K\subset L$, we have (by a straightforward linear algebra argument), with hopefully self-explanatory notation $$\dim_K \mathrm{InvBil}_K(\mathfrak{g})=\dim_L \mathrm{InvBil}_L(\mathfrak{g}\otimes_KL).$$

Consequence: for $\mathfrak{g}$ semisimple and $K$ of characteristic zero, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})$ is equal to the number of simple factors of the "complexification" $\mathfrak{g}\otimes_K\bar{K}$.

In particular, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})=1$ if and only if $\mathfrak{g}$ is absolutely simple.

Hence, to get counterexamples, it is enough to exhibit simple Lie algebras that are not absolutely simple. If $L$ is a finite extension of $K$ of degree $\ge 2$ and $n\ge 2$, $\mathfrak{sl}_n(L)$, viewed as Lie algebra over $K$, is such a Lie algebra.