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.

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 $-K_C$), 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.

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 $\mathbb{Q}$.

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.