Let $X$ be an algebraic variety defined over a field $k$ of characteristic zero. Suppose $X(K)$ is non-empty for some extension $K/k$. Is it true that $X(L)$ is non-empty where $L$ is the algebraic closure of $k$ in $K$? An equivalent formulation is: suppose $X(k)$ is empty; does that force $X(K)$ to be empty for every purely transcendental extension $K/k$?
In my particular application $k$ is a number field with a real place and $K=\mathbb{R}$ and then claim is true since the theory of real closed fields eliminates quantifiers, but I'm wondering about the general case. Relatedly Chevalley's Theorem (elimination of quantifiers for algebraically closed fields) does produce $\bar{k}$-points, but I don't see why they have to be $\bar{k}\cap K$-points.