If you're taking the definition of rational to be: birational to $\mathbb{P}^1$ over the field $k$, then the stated property is not even true. There are conics which have no rational points, and so are not rational, but are rational over a quadratic extension. For example, the affine conic $x^2 + y^2 + 1 = 0$ over the field $\mathbb{Q}$.

Added: since you only need the result when $L/k$ is a pure trancendental extension - there the result is indeed true. The idea is that you can represent any genus $0$ curve as a conic (take the embedding corresponding to $-2K_C$), and so you need to show that if the conic has a rational point over a pure transcendental extension of $k$, then it has a rational point over $k$ itself. There are many ways to see this (essentially you specialize the indeterminates suitably so that no denominators vanish) - one way is by induction on the transcendence degree. For a reference see Lam's "Quadratic forms over fields", Lemma 1.1 of Chapter IX. The result holds more generally for quadratic forms in any number of variables.